Question

Easy Question....!

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)


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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

$\large{\underline{\underline{\textbf{\textsf{Factorisation of \bf{32x^{8}-2 y^{8}} }}}}}$

$\sf{32x^8-2y^8}$

$\bigstar\textbf{\textsf{ Rewrite 32 as 16 \times 2}}$

$\sf{=16\cdot \:2x^8-2y^8}$

$\bigstar\textbf{\textsf{ Factor out Common Term 2}}$

$\sf{=2\left(16x^8-y^8\right)}$

$\rule{315}{1}$

$\bigstar\textbf{\textsf{ Factor 16 ^8 - y ^8 }}$

$\sf{=\left(4x^4\right)^2-\left(y^4\right)^2}$

$\sf{=\left(4x^4+y^4\right)\left(4x^4-y^4\right)}$

$\sf{=\left(4x^4+y^4\right)\left(2x^2+y^2\right)\left(2x^2-y^2\right)}$

$\rule{315}{1}$

$\sf{=2\left(4x^4+y^4\right)\left(2x^2+y^2\right)\left(2x^2-y^2\right)}$

$\large{\underline{\underline{\textbf{\textsf{ \therefore Option 2 is The Right Answer}}}}}$