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Question Number 105023 by bemath last updated on 25/Jul/20
(x2−y2)2(1−x2)(y2−1)+(1−x2)2(x2−y2)(y2−1)+(y2−1)2(1−x2)(x2−y2)=
Answered by john santu last updated on 25/Jul/20
(x2−y2)3+(1−x2)3+(y2−1)3(1−x2)(y2−1)(x2−y2)[set(x2−y2)=u;(1−x2)=v;(y2−1)=w]⇔u3+v2+w3uvw=(u+w)3−3uw(u+w)+v3uvw[u+w=x2−1=−v]⇒−v3+3uvw+v3uvw=3uvwuvw=3(JS★♠)
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