Question

Factorize a power 7 - 64a​

Answer 1

Step-by-step explanation:

a^7 - 64a

= a(a^6 - 64)

= a{(a³)² - (2³)²}

= a{(a³+2³)(a³-2³)

= a[{(a+2)(a²-2a+4)}{(a-2)(a²+2a+4)}]

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Answer 2

a⁷ - 64a = a(a + 2)(a - 2)(a² - 2a + 4)(a² + 2a + 4)

Given :

The expression a⁷ - 64a

To find :

To factorise the expression

Formula :

• a² - b² = ( a + b) ( a - b)

• a³ + b³ = (a + b)(a² – ab + b² )

• a³ - b³ = (a - b)(a² + ab + b² )

Solution :

Step 1 of 2 :

Write down the given expression

The given expression is

a⁷ - 64a

Step 2 of 2 :

Factorise the expression

$\sf {a}^{7} - 64a$

$\sf = a({a}^{6} - 64)$

$\sf = a \bigg[ {( {a}^{3} )}^{2} - {(8)}^{2} \bigg]$

$\sf = a( {a}^{3} + 8 )( {a}^{3} - 8 )$

$\sf = a( {a}^{3} + {2}^{3} )( {a}^{3} - {2}^{3} )$

$\displaystyle \sf{ =a\bigg[(a + 2)( {a}^{2} - a.2 + {2}^{2} ) \bigg] \bigg[(a - 2)( {a}^{2} + a.2 + {2}^{2} ) \bigg] }$

$\displaystyle \sf{ =a\bigg[(a + 2)( {a}^{2} - 2a + 4) \bigg] \bigg[(a - 2)( {a}^{2} + 2a + 4) \bigg] }$

$\displaystyle \sf{ =a(a + 2)(a - 2)( {a}^{2} - 2a + 4)( {a}^{2} + 2a + 4) }$

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