Factorise :-
$a^{4} - 2 a^{2} b^{2} + b^{4}$

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Solution

$a^{4} - 2 a^{2} b^{2} + b^{4} = a^{4} - a^{2} b^{2} - a^{2} b^{2} + b^{4} = a^{2} (a^{2} - b^{2}) - b^{2} (a^{2} - b^{2}) = (a^{2} - b^{2})^{2}$

As we know that

$x^{2} - y^{2} = (x + y) (x - y)$

$⟹ a^{4} - 2 a^{2} b^{2} + b^{4} = (a^{2} - b^{2})^{2} = (a - b)^{2} (a + b)^{2}$

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Similar questions

If X = 1 - a^{2} and Y = 1 - b^{2}, then what is X^{2} + Y^{2} ?

△ A B C

2 a^{2} b^{2} + 2 b^{2} c^{2} = a^{4} + b^{4} + c^{4}, ∠ B =

(i) a^{4} - b^{4} (i i) p^{4} - 81 (i i i) x^{4} - {(y + z)}^{4}

(i v) x^{4} - {(x - z)}^{4} (v) a^{4} - 2 a^{2} b^{2} + b^{4}

△ A B C, a^{4} + b^{4} + c^{4} = 2 a^{2} b^{2} + b^{2} c^{2} + 2 c^{2} a^{2},

sin A

a^{4} - b^{4}

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