Question

x^{​​​​​​6} - 729 (Factorise)

Answer 1

Rewrite x6 as (x2)3
(x2)3−729(x2)3-729
Rewrite 729 as 9Cube
(x2)3−9cube(x2)3-9cube

Since both terms are perfect cubes, factorusing the difference of cubes formula,
a3−b3=(a−b)(a2+ab+b2)a3-b3 where a=x2a=x2 and b=9b=9.
(x2−9)((x2)2+x2⋅9+9square)

Rewrite 9 as 3square
(x2−3square)((x2)2+x2⋅9+9square)
Since both terms are perfect squares, factor using the difference of squaresformula, a2−b2=(a+b)(a−b)a2-b2=(a+b)(a-b)where a=xa=x and b=3b=3.
(x+3)(x−3)((x2)2+x2⋅9+9square)
Multiply the exponents in (x2)2.

(x+3)(x−3)(x4+x2⋅9+9square)

Move 9 to the left of the expression x2⋅9x2⋅9.
(x+3)(x−3)(x4+9⋅x2+9square)
Multiply 9 by x2 to get 9x2
(x+3)(x−3)(x4+9x2+9square)


Raise 9 to the power of 2 to get 81.
(x+3)(x−3)(x4+9x2+81)

Rewrite the middle term.
(x+3)(x−3)(x4+(2x2⋅9−9x2)+81)


Rearrange terms.
(x+3)(x−3)(x4+2x2⋅9+81−9x2)
Factor first three terms by perfect square rule.
(x+3)(x−3)((x2+9)2−9x2)

Rewrite 9x2 as (3x)2.
(x+3)(x−3)((x2+9)2−(3x)2))
Since both terms are perfect squares, factor using the difference of squaresformula, a2−b2=(a+b)(a−b)a2-b2=(a+b)(a-b)where a=x2+9a=x2+9 and b=3xb=3x.
(x+3)(x−3)((x2+9+3x)(x2+9−(3x))

Multiply 3 by -1 to get −3
(x+3)(x−3)((x2+9+3x)(x2+9−3x))(x+3)(x-3)((x2+9+3x)(x2+9-3x))
Remove unnecessary parentheses.
(x+3)(x−3)(x2+9+3x)(x2+9−3x)