Historian Office.

Salt Lake City.

Utah Territory.

Sept. 2, 1876

Dr. A. B. Nelson.

Danville Ky.

Dear Sir:

Yours of the 1^{st} ult. was duly received. Your candid criticisms upon my general forms of the roots of the cubic equation, I highly appreciate. The forms of (A), although applicable to every possible variety of imaginary roots, were only intended as general expressions for real roots.

In the general equation, [marked in my mfs. (1),] you assume q = 0 : the resulting binomial contains two imaginary roots. My forms (marked (A)) give the following values for x:

± r^{1/3} , ± r^{1/3} /2 ± r^{1/3} /2 √(-3) , ± r^{1/3} /2 ± r^{1/3} /2 √(-3) .

In this case, you proved –m to be, not only a real positive, but a real quantity: this is correct; hence, +m, in each of the last terms of the last two forms of (A), is necessarily rendered imaginary, as it should be in giving the two imaginary roots.

You express a belief that “m is Cardan’s irreducible radical.” But it must be remembered that the two cubes of Cardan are

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2

only irreducible, when the roots are real. Whereas in the forms of (A), m is always real, when the roots are real, and vice versa, m is only imaginary, when the roots are imaginary. I shall now proceed to prove, by another process, that m is a real quantity – that it is equal to the one-twelfth part of the square of one of the differences between the roots of (A). Subtract the second from the third form, and we have

2√(3m) = diff. = z;

hence

m = z²/12 .

Now the difference between real roots is real; therefore, z, z² and m must be real: and <m> is not, as you have expressed, “a cubic radical.”

That m, and all other real quantities, can, when <each is> combined with other real quantities in a cubic form, be thrown into the shape of Cardan’s formula, is not for a moment disputed.

For instance, in the forms of (A), <using the lower signs,> subtract the third from the first, and we have

+3 ( q/3 – m)^{½} – (√(3m)

Again subtract the first from the second, and we have

– 3 ( q/3 – m)^{½} – (√(3m)

Subtract the second from the third, and, (as above,) we have

+ 2√(3m)

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3

By these three differences, we form the general equation of differences, namely,

z^{3} – 3qz + 24√(3m) (m^{3/2} – (q/4)m^{½}) = 0 ……………….. (d)

But

24√3 (m^{3/2} – (q/4) m^{½}) = ± √(4q³-27r²) = ± r¹

q in this last formula must be considered positive.

Divide the roots of <(d)> by 2√3 , and we have

z’³ – (q/4) z’ + (m^{3/2} – (q/4)m^{½}) = 0 ………………………..(e)

The roots of (e) are

+√3/2 (q/3 – m)^{½} – √m/2 , +√m , -√3/2 (q/3 – m)^{½} – √m/2

But the final term of (e) equals a known quantity, namely,

m ^{3/2} -(q/4) m^{½} = ±(√(4q³-27r²)/24√3) = ± r’/24√3

In this equation let q/4= q’’ , and ± r’/24√3 = ± r”, and we have

m^{3/2} – q”m^{1/2} ± r” =0.

Applying Cardan’s formula, we have

m^{½} = [r”/2 + (r”²/4 +q”³/27 )^{½}]^{¹/³} + [- r”/2 – (r”²/4 + -q³/27 )^{½}] ^{1/3}

If q = 0, then q” = 0, and we have

m^{½} = – r”^{1/3};

hence

m = r’’ ^{2/3} =r’^{2/3}/(24√3)^{2/3} = r^{2/3}/12 = (-27r²)^{1/3}/12= (-1/4) r²/³.

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4

The other two values of m are imaginary, as they should be when q = 0. But these values, by the forms of (A), are never imaginary, when q is of such a value that the three roots are real. But by Cardan’s formula they are obliged to assume the imaginary form, the same as all other real quantities must do, if they enter as roots into the cubic form, and are subjected to Cardan’s process.

There is evidently some error in your formula which you sent to me, representing the value of m. You find

m= ³√(r²/32 – q³/216 – r/4 √(r²/64 – q³/216)

By assuming q = 0, you inadvertently make m= 1/4 ³√r².

By a review, you will find m= -1/³√16 · ³√r² = 1/2³√2 ³√r².

You will undoubtedly find the error in the formula: it should be such as to give the same value of m as you have given, and as I have also given on the preceding page.

You have very candidly arrived at the conclusions that my formula possesses no superiority over that of Cardan’s. I hope you will not consider me egotistical, when I advance a few reasons, showing its superiority.

First, Cardan’s formula, though expressed in known

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Required the General Form of the Roots of the Cubic Equation

x³ – qx r = 0 …………………………….(1)

Roots

±2(q/3 – m)^{1/2 }, ±[(q/3 – m)^{1/2 }-√3m], ±[(q/3 – m)^{1/2 }+√3m].

The sum of the products of these roots, taken two and two, is equal to q. Their continued product, with the signs changed, is equal to ±2[(q/3 – m)^{3/2}-3m(q/3 – m)^{1/2} ] =±r.

In these roots, if m were known, we should have the general solution: but m is unknown; therefore, the roots are only general in form. The simplicity of this form, is far superior to Cardan’s Irreducible Form. An infinite number of equations can be solved for each given value of m.

When m=q/12, one of the roots is nothing: hence, the final term becomes nothing: the two remaining roots equal ±√q.

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By Cardan’s formula, we have

x =( r/2)^{1/3} { [ 1 + (y-1)^{½} ] ^{1/3}+ [1 – (y-1)^{½} ]^{1/3}}, in which y = q³/6.75r² .

Let [1 + (y-1)^{½} ]^{1/3} = u + v √-1 ; it is required to find v in terms of u. We have

(u + v√-1)³ = 1 + (y-1)^{½} √-1 = u³ – 3v²u + (3 vu² – v³) But u³ = 3uv² = 1; therefore

v² = (u³-1)/3u = 1/3 (u² – 1/u ); hence

v = ± (1/√3) 1 (u² – 1/u )^{½}; therefore

u ± v√-1 = u ± (1/√3) (u² – 1/u )^{½ } √-1

We also have

(3v u^{2} – v^{3}) √-1 = v (3u² – v³)√-1 = v (3 u² – v²) √-1 . By substituting the value of v and v², we have

1/√3 (u² – 1/u )^{½} [3u² – 1/3 (u² – 1/u)] = (y – 1)^{½
}=1/√3 (u² – 1/u)^{½} 1/3(8u² + 1/u)

Therefore we have

[1 + 1/√3 (u² – 1/u)^{½} 1/3 (8u² + 1/u)√-1 ]^{1/3} = u + 1/√3 (u² – 1/u)^{½} √-1

= [1 + (y -1)^{½} √-1 ]^{1/3}

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Given x^{3} – (5×2^{1/3})/3^{1/3} . x -2 = 0, to find its roots.

Hence

x = 1/18 (1 + 1 4/9 √-1 )^{1/3}. By trial, it will be found that

[1/18^{1/3} (3 + √-1 )]³ = 1/18 (1 + 1 4/9 √-1); therefore the positive root of the given equation will be equal to 6/18^{1/3}= 2^{2/3} x 3^{1/3} , and therefore the two negative roots will be

(-3 +√3)/18^{1/3}, (-3 -√3)/18^{1/3}

Given x³ – 3 (4913/2704 )^{1/3} x – 2 = 0 , to find its roots.

Let q = 3 [((3n² -1)/(n³-3n))² + 1]^{1/3} . In this n = 4; and therefore in Cordan’s formula, the two cube roots required will be

1/(n³-3n)^{1/3} (n +√-1), 1/(n³-3n)^{1/3} (n -√-1

Or

1/52^{1/3} (4 + √-1), 1/52^{1/3} (4 – √-1) ; and therefore, one root will be

x = 8/52^{1/3} =(2 x 2^{2/3} x 2^{2/3} )/18^{1/3}; the two negative roots will be

x = – 4/52^{1/3} + 1/52^{1/3} √3 ; x = – x = 1 4/52^{1/3} – 1/52^{1/3} √3 .

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Required the three roots of equation

x^{3} – 3 [((3n²-1)/(n³-3n))² +1]^{1/3}. (r/2)^{2/3 }. x – r = o ,

in terms of n and r.

x = 2n/((n³-3n)^{1/3}) · (r/2 )^{1/3}.

x = (-n+√3)/((n³-3n)^{1/3}) · (r/2 )^{1/3}

x = (-n-√3)/((n³-3n)^{1/3}) · (r/2 )^{1/3}

And for the three roots of the equation

x³ – 3 [ ((3n²-1)/(n³-3n))² + 1]^{1/3} . (r/2)^{2/3} x + r = 0, we have the above roots with the signs changed

x = -2n/((n³-3n)^{1/3}). (r/2)^{1/3}

x = (n-√3)/((n³-3n)^{1/3 }) . (r/2)^{1/3}

x = (n+√3)/((n³-3n)^{1/3} ). (r/2)^{1/3}

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~~The first factor in the second coefficient can be reduced to a simpler form: thus~~

~~ x³ – 3 (n²+1)/((n³-3n)~~^{2/3}), (r/x)^{2/3} x ± r = 0.

~~The roots of this will be the same as those on the preceeding page.~~

~~In the equation, x³ – q x ± r = 0 (1)~~

~~let q = a (r/2)~~^{2/3} : and let a/3 = (n²+1)/((n³-3n)^{2/3} ). In these equations, q, r, a and (n²+1)/((n³-3n)^{2/3} ) are all known quanties; while n is any number

~~The following are the six roots of equation (1):~~

~~ x = ±2√a/√3 . n/((n+1)~~^{1/3}) . (r/2)^{1/3}

~~ x = ± √a/√3 . (n-√3)/((n²+1)~~^{1/2})·(r/2)^{1/3}

~~ x = ± √a/√3 . (n+√3)/((n²+1)~~^{1/2})·(r/2)^{1/3}

~~For the upper sign of r in equation (1), the roots with upper signs must be taken; and vice versa.~~

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1

General Forms of the Roots.

x³ – qx ± r = 0………………………………………………(1)

x = ±2 (q/3 – m)^{1/2} ………………………………………………..(a)

x = ± (q/3 – m)^{1/2} ±√(3m) …………………………………………(b)

x = ± (q/3 – m)^{1/2} ±√(3m)………………………………………….(c)

The resulting equation, equivalent to equation (1), will be

x³ – qx ±2 [(q/3 – m)^{3/2} – 3m (q/3 – m)^{1/2} ] = 0 ……………………(2)

When – m = 0, equation (2) becomes

x³ – qx ± 2 (q^{2/3}/3√3) = 0 …………………………………………………(3)

For the roots of this, the forms (a), (b) and (c) give

X = ± 2(√q/√3) ; x =± (√q/√3) ; x =(√q/√3) .

When m = q/3 , equation (2) becomes x³ – qx = 0; o x and (a), (b) and (c) give

x = 0; x = ±√q ; x =±√q .

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2

When m = x =q/12 , the two roots represented by (b) = 0 while (a) and (c) will give x = ±√q , x =±√q

When – m is negative and <numerically> greater than q/3 , the roots, represented by (a), (b) and (c), will all be imaginary, and the compound final term of equation (2) will be imaginary in both of its simple terms.

When – m is negative and numerically less than , the roots (a), (b) and (c) will be real, except in the particular cases case where (a) or (b) becomes nothing.

When –m is positive the root (a) will be real, but the roots (b) and (c) will be imaginary.

In all the foregoing cases q is supposed positive, except in (1). If q in (1) is positive, it must <be> considered negative in the following.

If –q and –m are both negative, all the roots of (a) (b) and (c) will be imaginary.

If – m is positive and –q negative, and if – q/3 is numerically greater than –m then (a) will be imaginary, while (b) and (c) will each have its two terms imaginary. But if – q/3 is numerically less than the positive quantity – m, (a) will be real, but (b) and (c) will each have its last term imaginary.

If – q/3 is equal to the positive quantity –m, then (a) will be zero, while (b) and (c) will become ±√-q , ±√-q .

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3

If q = 0, and –m negative, the final terms in equations (1) and (2) will be imaginary, and the roots (a), (b) and (c) will all be imaginary.

If q = 0, and –m positive, the final terms in equations (1) and (2) will be real; also (a) will be real, but (b) and (c) will be imaginary. In this case, we have

x = ±r^{1/3}, x = ± 2√-m , – m being positive;

hence

-m = +(1/4) r ^{2/3}.

If this value of m be substituted in equation (2), it will reduce to x³ ± r = 0. Also if the same value of m be substituted in (a), (b) and (c), the roots will become

±r ^{1/3} , ± r ^{1/3}/2 ± (r ^{1/3}/2)√-3 , ±r ^{1/3}/2 ± (r ^{1/3}/2)√-3.

The sum of the products of these roots, taken two and two, is equal to nothing; their continued product with the signs changed, is equal to ± r.

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4

Equation of differences

z^{3} – 3qz + 2[(3m)^{3/2 }– 9 (q/3 – m) (3m)^{½ }] = 0.

Or z^{3} – 3qz + 24√3 (m^{3/2 }– (q/4)m^{½}) = 0……………………………………………..(4)

But 24√3 (m^{3/2 }-(q/4) m^{½}) = √(4q³-27r²) = r^{’}, q being positive.

Roots of (4)

±3 (q/3 -m)^{½ }±√3m , ±2√3m , ± 3 ((q/3 – m)^{½} ±√3m .

Divide the roots of (4) by 2√3m , and we have

z^{’ 3} –(q/4) z^{’ }+ (-m ^{3/2} +(q/4) m^{1/2}) = 0……………….…………….(5)

But

-m ^{3/2} +(q/4) m^{½ }= + √(4q³-27r²)/(24√3) = r’/(24√3) …………………….……………..(6)

Roots of (5)

±√3/2 (q/3 – m)^{½} ±√m/2 , ±√m , ± √3/2 (q/3 – m)^{½} ±√m/2 .

By Cardan’s formula Let q/4= q^{’’}; and = r^{’’}/24√3 = r”,

m^{3} = q Then equation (6) becomes

m^{3/2} – q’’ m^{½} ± r’’ = 0…………………………………………… (7)

A root of (7), by Cardan’s formula, becomes

m^{½} = [- +r”/2 + (r”²/4 + -q”³/27 )^{½}]^{1/3} + [ – +r”/2 – (r”²/4 + -q”³/27)^{½}]^{1/3} ……..(d)

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5

When <If> q=0 < then> q” = 0, and formula (d) becomes

m ^{1/2} = r” ^{1/3}

hence m = r”^{2/3} = r’^{2/3}/(24√3)^{2/3} = r’^{2/3}/12 = (27r²)^{1/3}/12 = -1/4 r ^{2/3}

[n + ((r+2n³)/2.3n )^{1/2} √-1]^{3 } = – r/2 + √((r²/4) + (q³/27)).

If in the general forms of (A), m = q/4, two of the roots will become equal, and their value will be as follows:

±√q/√3 , ±√q/√3 , ± 2√q/√3.

If in equation (5), m = , <one> of the roots will reduce to nothing, the remaining roots will be ±√q/2 , ±√q/√3 .

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6.- x^{4} – 1 =0

The roots of these two factors are the same as those in examples 1 and 2.

7.- x^{3} + 1 = (x^{4} – √-1 ) (x^{4} + √-1 ) = 0

Find a root of x^{4} = + √-1 . Answer x = ((√2 +1)/(2√2))^{½} + ((√2 +1)/(2√2))^{½ }√-1

Find a root of x^{4} = – √-1 . Answer x = ((√2 +1)/(2√2))^{½} – ((√2 +1)/(2√2))^{½ }√-1

x² – √-1 = (x – (1/√2) – (1/√2) √-1 ) (x + (1/√2) + (1/√2) √-1 ) = 0

x² + √-1 = (x – (1/√2) + (1/√2) √-1 ) (x + (1/√2) – (1/√2) √-1 ) = 0

x^{4} – √-1 = [x² – (1/√2) + (1/√2) √-1] [x² +(1/√2) + (1/√2) √-1] = 0

But

x² – ( (1/√2) + (1/√2) √-1 = [x + ((√2 +1)/(2√2))^{½} + ((√2 -1)/(2√2))^{½} √-1 ] [x – ((√2 +1)/(2√2))^{½} – ((√2 -1)/(2√2))^{½} √-1 ] = 0;

and

x² + ( (1/√2) + (1/√2) √-1 = [x + ((√2 +1)/(2√2))^{½} – ((√2 -1)/(2√2))^{½} √-1 ] [x – ((√2 +1)/(2√2))^{½} + ((√2 -1)/(2√2))^{½} √-1 ] = 0;

x^{4 }+ √-1 = (x^{2 }– (1/√2)+ (1/√2) √-1) (x^{2 }+ (1/√2) – (1/√2) √-1) =0

But

x^{2 –} (1/√2)+ (1/√2) √-1 = [x -((√2 +1)/(2√2))^{½} +((√2 -1)/(2√2)^{½} √-1 ] [x +((√2 +1)/(2√2))^{½} -((√2 -1)/(2√2)^{½} √-1 ] =0;

and

x^{2 } +(1/√2)- (1/√2) √-1 = [x +((√2 +1)/(2√2))^{½} +((√2 -1)/(2√2)^{½} √-1 ] [x -((√2 -1)/(2√2))^{½} -((√2 -1)/(2√2)^{½} √-1 ] =0;

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x^{16 }+ 1 = 0 = (x^{8} – √-1) (x^{8 }+ √-1 ) .

x^{8 }– √-1 = [x^{4} – (√-1) ^{1/2} ] [x^{4} + ( √-1 )^{1/2}] ; x^{8} + (√-1 = {x^{4} – ( -√-1 )^{1/2}} {x^{4} + (- √-1 )^{1/2} } = 0.

x^{4} – (√-1 )^{1/2} = [x² – (√-1 )^{1/4} ] [x² + (√-1 )^{1/4} ] = 0

x^{4} + (√-1 )^{1/2} = [x² – (√-1 )^{1/2} ] [x² + (√-(√-1))^{1/2} ] = 0

x^{4} – √-(√-1) = [x² – ( – √-1 ^{1/4}] [x² + ( – √-1 ^{1/4}] = 0

~~x~~^{4} – √-(√-1) = [x² – ( – √-1 ^{1/4}] [x² + ( – √-1 ^{1/4}] = 0

x^{4} + ( – √-1 ^{1/2} = {x² – [ – ( – √-1) ^{1/2}]^{½}} {x² + [ – (- √-1 ) ^{1/2}]^{½}} = 0

x² – (√-1 )^{¼} = [x – (√-1) ^{1/8} ] [x + ( √-1) ^{1/8} ] = 0

x² + √-1 )^{¼} = {x – [ – (√-1 ) ^{1/4} ]^{½}} {x + [- ( √-1)^{1/4} ]^{½}} = 0

x² – [ – (√-1 ) ^{1/2} ]^{½} = {x – [ – (√-1) ^{1/2} ]^{¼}} {x + [ – (√-1 ) ^{1/2} ]^{¼} }

x² + [ – (√-1) ^{1/2} ]^{½} = {x – [[- [- ( √-1)^{1/2} ]^{½}]]^{½}} {x + [[-[ – ( √-1)^{1/2} ]^{½}]]^{½}} = 0

x² – (- √-1 )^{¼} = [x – ( – √-1)^{1/8} ] [x + ( -√-1) ^{1/8} ] = 0

x² + (- √-1 )^{¼} = {x – [- ( – √-1) ^{1/4} ]^{½}} { x + [- ( -√-1) ^{1/4} ]^{½}} = 0

x² – [- (- √-1 )^{½}]^{½} = {x – [- ( -√-1) ^{1/2} ]^{¼}} { x + [- ( -√-1) ^{1/2} ]^{¼}} = 0

x² +[- (- √-1 )^{½}]^{½} = {x – [[- [- ( -√-1) ^{1/2} ]^{½}]]^{½}} {x + [[- [- ( -√-1) ^{1/2} ]^{½}]]^{½}} = 0

x² – (√-1)^{¼} = {x + [1/2 + 1/2 (1/2 +1/2 . 1/√2)^{½}]^{½} + [1/2 -1/2 (1/2 +1/2 . 1/√2 )^{½}]^{½} √-1 }x

{x – [1/2 + 1/2 (1/2 +1/2 . 1/√2 )^{½}]^{½} – [1/2 – 1/2 (1/2 + 1/2. 1/√2 )^{½}]^{½} √-1 }= 0

x² + ( √-1 )^{¼} = {x + [1/2 – 1/2 (1/2 + 1/2 . 1/√2 )^{½}]^{½} – [ 1/2 + 1/2 (1/2 + 1/2 . 1/√2 )^{½}]^{½} √-1}x

{x – [1/2 – 1/2 (1/2+ 1/2 . 1/√2 )^{½}]^{½} + [1/2 +1/2 (1/2 + 1/2 . 1/√2 )^{½}]^{½} √-1 }= 0

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1.– Find the four biquadratic roots of the equation

Roots.

x^{4 }+ C = 0. x^{4 }+ 1 = 0

x = C^{1/4} (-1)^{1/4}^{ }; x = 1/√2 (+ √-1 )

x = –C^{1/4}(-1)^{1/4} ; x = 1/√2(1- √-1 )

x = C^{1/4 }(-1)^{3/4} ; x = 1/√2(-1 + √-1 )

x = -C^{1/4} (-1)^{3/4} . x = 1/√2 (-1 – √-1 )

2. – Find the four biquadratic roots of the equation

Roots. x^{4} – C = 0 . x^{4} – 1 = 0 .

x = C^{1/4 } ; x = +1 ;

x = -C^{1/4} ; x = – 1 ;

x = C^{1/4} (-1)^{1/2} ; x = + √-1

x = -C^{1/4 }(-1)^{1/2} . x = – √-1

3. – Find the seven roots of unity in the equation

x^{7 }– 1 = 0 .

x = 1 .

(x^{7}-1)/(x-1) = x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x +1 = 0; divide by x^{3}, and we have

x³ +(1/x³) + x² +(1/x²) + x +(1/x) + 1 = 0. Put

Put x + (1/x) = z ; hence x² + (1/x²) = z² – 2 ; and

x³ + (1/x³) = z³ – 3z; therefore

x³ +(1/x³) + x² + (1/x²) + x+ (1/x) + 1 = z³ + z² – 2z – 1 = 0 ; hence

z = 1.246970186

z = 1.801949479

z = .445020707

We also have x² – zx + 1 = 0 ; substitute the three values of z

(x² – 1.246970186 x + 1) (x² + 1.801949479 x+1) (x² + .445020707 x+1)

= x^{6} + x^{5} + x^{4} + x ³ + x² + x + 1 = 0. The roots of these quadratics will give the remaining six roots of the equation x^{7} – 1 = 0.

[page break]

(z³+z²-2z-1)/(z-1.246970186) = z² + 2.246970186 z +.801904831 = 0.

z = -1.123485093 ± √(1.123485093)² – .801904831)

or z = – 1.123485093 ± .678464386

4. – x^{5} – 1 = 0.

x = 1.

x = 1/4 [- (1 – √5) + √2 √(-5-√5) ] ;

x = 1/4 [- (1 – √5) – √2 √(-5-√5) ] ;

x = 1/4 [- (1 – √5) + √2 √(-5 + √5) ] ;

x = 1/4 [- (1 – √5) – √2 √(-5 + √5) ] ;

These five roots with their signs changed will be the five roots of the equation x^{5} + 1 = 0.

5. – x^{6} – 1 = 0.

x = + 1 ;

x = -1 ;

x = +( 1/√2) (√(-1 + √-3))

x = +( 1/√2) (√(-1 – √-3))

x = +( 1/√2) (√(-1 + √-3))

x = +( 1/√2) (√(-1 – √-3))

x^{6} + 1 = 0.

x = + √-1 ;

x = -√-1 ;

x = +( 1/√2) (√(1 + √-3) ;

x = +( 1/√2) (√(1 – √-3) ;

x = +( 1/√2) (√(1 + √-3) ;

x = +( 1/√2) (√(1 – √-3) ;

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