(x^4 )^2+(y^4 )^2 =(x^4+y^4 )^2 2x^4 y^4 =(x^2 )^2+(y^2 )^2 ^2 2x^4 y^4 =(x^2+y^2 )^2 2x^2 y^2 ^2 2x^4 y^4 =[{(x^2+y^2 √(2) xy)(x^2+y^2+√(2) xy)} √(2) x ...

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Byju's Answer

Standard IX

Mathematics

(a ± b)²

Factorise x8+...

Question

Factorise $x^{8} - y^{8}$

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Solution

$(x^{4})^{2} - (y^{4})^{2}$
$= (x^{4} - y^{4}) (x^{4} + y^{4}) [Using identity a^{2} - b^{2} = (a - b) (a + b)]$
$= {(x^{2})^{2} - (y^{2})^{2}} (x^{4} + y^{4})$
$= {(x^{2} - y^{2}) (x^{2} + y^{2})} (x^{4} + y^{4})$
$= (x - y) (x + y) (x^{2} + y^{2})} (x^{4} + y^{4})$

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Factorise x8−y8

(x4)2−(y4)2=(x4−y4)(x4+y4) [Using identitya2−b2=(a−b)(a+b)]={(x2)2−(y2)2}(x4+y4)={(x2−y2)(x2+y2)}(x4+y4)=(x−y)(x+y)(x2+y2)}(x4+y4)

Factorise $x^{8} - y^{8}$

$(x^{4})^{2} - (y^{4})^{2}$
$= (x^{4} - y^{4}) (x^{4} + y^{4}) [Using identity a^{2} - b^{2} = (a - b) (a + b)]$
$= {(x^{2})^{2} - (y^{2})^{2}} (x^{4} + y^{4})$
$= {(x^{2} - y^{2}) (x^{2} + y^{2})} (x^{4} + y^{4})$
$= (x - y) (x + y) (x^{2} + y^{2})} (x^{4} + y^{4})$