Question

What is the complete factorisation of 32x^8 - 2y^8?
(a) 2(4x^2 + y^2)(4x^2 - y^2)
(c) (4x^2 + y^2)(8x^2 - 2y^2)
(b) 2(2x^2 + y^2)(2x^2 - y^2)(4x^4 + y^4)
(d) (4x + 2y)(2x - y)(4x^2 + y^2)
​

Answer 1

Solution :

To factorise :

> 32 x⁸ - 2y⁸

> 2 [ 16 x⁸ - y⁸ ]

> 2 [ ( 4x⁴ )² - ( y⁴ )² ]

> 2 [ 4x⁴ + y⁴ ][ 4x⁴ - y⁴ ]

> 2 [ 4x⁴ + y⁴ ][ ( 2x² )² - ( y² )² ]

> 2 [ 4x⁴ + y⁴ ][ 2x² + y² ][ 2x² - y² ]

Hence , Option ( b) is the correct answer .

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Additional Information :

(a + b)² = a² + 2ab + b²

(a + b)² = (a - b)² + 4ab

(a - b)² = a² - 2ab + b²

(a - b)² = (a + b)² - 4ab

a² + b² = (a + b)² - 2ab

a² + b² = (a - b)² + 2ab

2 (a² + b²) = (a + b)² + (a - b)²

4ab = (a + b)² - (a - b)²

ab = {(a + b)/2}² - {(a-b)/2}²

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

(a + b)³ = a³ + 3a²b + 3ab² b³

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)( a² - ab + b² )

a³ + b³ = (a + b)³ - 3ab( a + b)

a³ - b³ = (a - b)( a² + ab + b²)

a³ - b³ = (a - b)³ + 3ab ( a - b )

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Answer 2

${\Large{\bold{\rm{\red{Question}}}}} \; \; \; \; \ddag$

What is the complete factorisation of 32x⁸ - 2y⁸ ?

(a) 2(4x² + y²)(4x² - y²)

(b) 2(2x² + y²)(2x² - y²)(4x⁴ + y⁴)

(c) (4x² + y²)(8x² - 2y²)

(d) (4x + 2y)(2x - y)(4x² + y²)

${\Large{\bold{\rm{\red{Required \; Solution}}}}} \; \; \; \; \ddag$

${\tt{\underline{Let's \: factorise \: the \: given \: term \: completely}}}$

${\bold{\sf{\pink{32x^{8} - 2y^{8} \; \; \; \green \bigstar}}}}$

${\bold{\sf{:\implies 2(16x^{8} - y^{8}}}}$

${\bold{\sf{:\implies 2[(4^{4})^{2} - (y^{4})^{2}]}}}$

${\bold{\sf{:\implies 2(4x^{4} + y^{4})(4x^{4} - y^{4}}}}$

${\bold{\sf{:\implies 2[4x^{4} + y^{4}][(2x^{2})^{2} - (y^{2})^{2}]}}}$

${\bold{\sf{:\implies 2(2x^{2} + y^{2})(2x^{2} - y^{2})(4x^{4} + y^{4}}}}$

${\small{\frak{Henceforth, \: option \: (b) \: 2(2x^{2} + y^{2})(2x^{2} - y^{2})(4x^{4} + y^{4} \: is \: correct \: answer}}}$

${\Large{\bold{\rm{\red{Additional \; information}}}}} \; \; \; \; \ddag$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\;b)^{2}\;=\;a^{2}\:+\:b^{2}\:+\:2ab}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:-\:b)^{2}\;=\;a^{2}\:+\:b^{2}\:-\:2ab}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\:b\:+\:c)^{2}\;=\;a^{2}\:+\:b^{2}\:+\:c^{2}\:+\:2ab\:+\:2bc\:+\:2ac}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\;b)^{3}\;=\;a^{3}\:+\:b^{3}\:+\:3ab(a\:+\:b)}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:-\;b)^{3}\;=\;a^{3}\:-\:b^{3}\:-\:3ab(a\:-\:b)}\end{gathered}$