Factorise 16 x^{4}-y^{4}. | vedclass

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Question

(A) To factorise $16 x^{4}-y^{4}$,we use the difference of squares identity: $a^{2}-b^{2}=(a-b)(a+b)$.First,rewrite the expression as $(4 x^{2})^{2}-(

Vedclass · Accepted Answer

$(4x^2 + y^2)(2x + y)(2x - y)$

Vedclass · Answer

(A) To factorise $16 x^{4}-y^{4}$,we use the difference of squares identity: $a^{2}-b^{2}=(a-b)(a+b)$.First,rewrite the expression as $(4 x^{2})^{2}-(y^{2})^{2}$.Applying the identity,we get $(4 x^{2}-y^{2})(4 x^{2}+y^{2})$.Now,factorise the term $(4 x^{2}-y^{2})$ further by writing it as $(2 x)^{2}-(y)^{2}$.Using the identity again,$(4 x^{2}-y^{2}) = (2 x-y)(2 x+y)$.Combining these,the final factorised form is $(2 x-y)(2 x+y)(4 x^{2}+y^{2})$.

Factorise $16 x^{4}-y^{4}$.

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