Question

factorise: x^6-7x^3-8​

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

x^6 - 7x³ - 8

★We have :-

★Assumption

x³ = y

= x^6 - 7x³ - 8 = y² - 7y - 8

= y² - 8y + y - 8

= y(y - 8) + (y - 8)

= (y - 8)(y + 1)

★Now :-

★y = x³

= (x³ - 8)(x³ + 1)

= [(x)³ - (2)³][(x)³ + (1)³]

= [(x - 2)(x² + 2x + 4)][(x + 1)(x² - x + 1)]

= (x - 2)(x + 1)(x² + 2x + 4)(x² - x + 1)

= x^6 - 7x³ - 8

= (x - 2)(x + 1)(x² + 2x + 4)(x² - x + 1)

$\boxed{\begin{minipage}{9 cm} Additional Information \\ \\ \ 1)(x+y)^2=x^2+y^2+2xy \\ \\ 2)(x-y)^2=x^2+y^2-2xy \\ \\ 3)(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx \\ \\ 4)(x+y)^3=x^3+y^3+3xy(x+y) \\ \\ 5)(x-y)^3=x^3-y^3-3xy(x-y) \end{minipage}}$

Answer 2

$\huge\boxed{SOLUTION}$

We have to factorise the given(x^6 - 7x³ - 8)

So,

$= x {}^{6} - 7 \times 3 - 8 = y {}^{2} - 7y - 8$

$= y {}^{2} - 8y + y - 8$

$= y(y - 8) + (y - 8)$

$= (y = 3 {}^{3} )$

$(y = 3 {}^{3} )$

=> (x³ - 8)(x³ + 1)

THEN

=> (x)³ - (2)³)((x)³ + (1)³)

=> ((x - 2)(x² + 2x + 4))((x + 1)(x² - x + 1))

=> (x - 2)(x + 1)(x² + 2x + 4)(x² - x + 1)

=> (x^6 - 7x³ - 8)

=> (x - 2)(x + 1)(x² + 2x + 4)(x² - x + 1)