Question
Factorise:
$x^6-729$
$x^6-729$
✓
Answer
$x^6-729$
$=\left(x^2\right)^3-(9)^3$
$=\left(x^2-9\right)\left[\left(x^2\right)^2+x^2 \times 9+(9)^2\right] \text { Since } a^3-b^3=(a-b)\left(a^2+a \times b+b^2\right)$
$=\left(x^2-9\right)\left(x^4+9 x^2+81\right)$
$=(x+3)(x-3)\left[\left(x^2+9\right)^2-(3 x)^2\right]$
$=(x+3)(x-3)\left(x^2+3 x+9\right)\left(x^2-3 x+9\right)$
$=\left(x^2\right)^3-(9)^3$
$=\left(x^2-9\right)\left[\left(x^2\right)^2+x^2 \times 9+(9)^2\right] \text { Since } a^3-b^3=(a-b)\left(a^2+a \times b+b^2\right)$
$=\left(x^2-9\right)\left(x^4+9 x^2+81\right)$
$=(x+3)(x-3)\left[\left(x^2+9\right)^2-(3 x)^2\right]$
$=(x+3)(x-3)\left(x^2+3 x+9\right)\left(x^2-3 x+9\right)$
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