Question

answer this........

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Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept\;:-}}}$

Here the Concept of Algebraic Identities had been used. We are given the equations. Firstly we will apply the identities and expand them. Then we will reduce them by taking the terms in common and hence, the simplification will be done.

Let's do it !!

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★ Identities Used :-

$\\\;\boxed{\sf{(x\;-\;y)^{2}\;=\;\bf{x^{2}\;+\;y^{2}\;-\;2xy}}}$

$\\\;\boxed{\sf{(x\;+\;y)^{2}\;=\;\bf{x^{2}\;+\;y^{2}\;+\;2xy}}}$

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★ Solution :-

1.] (a² - b²)²

2.] (ab + bc)² - 2ab²c

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1.) (a² - b²)² ::

Here we can apply the first identity that is,

$\\\;\rm{:\rightarrow\;\;(x\;-\;y)^{2}\;=\;\bf{x^{2}\;+\;y^{2}\;-\;2xy}}$

• Here x = a² and y = b²

$\\\;\sf{:\Longrightarrow\;\;(a^{2}\;-\;b^{2})^{2}\;=\;\bf{(a^{2})^{2}\;+\;(b^{2})^{2}\;-\;2(a^{2})(b^{2})}}$

$\\\;\bf{:\Longrightarrow\;\;(a^{2}\;-\;b^{2})^{2}\;=\;\bf{a^{4}\;+\;b^{4}\;-\;2a^{2}b^{2}}}$

$\\\;\underline{\boxed{\tt{(a^{2}\;-\;b^{2})^{2}\;=\;\bf{\blue{a^{4}\;+\;b^{4}\;-\;2a^{2}b^{2}}}}}}$

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2.) (ab + bc)² - 2ab²c

Here we can apply the second identity that is,

$\\\;\rm{:\rightarrow\;\;(x\;+\;y)^{2}\;=\;\bf{x^{2}\;+\;y^{2}\;+\;2xy}}$

• Here x = ab and y = bc

$\\\;\rm{:\Longrightarrow\;\;(ab\;+\;bc)^{2}\;-\;2ab^{2}c\;=\;\bf{(ab)^{2}\;+\;(bc)^{2}\;+\;2(ab)(bc)\;-\;2ab^{2}c}}$

$\\\;\rm{:\Longrightarrow\;\;(ab\;+\;bc)^{2}\;-\;2ab^{2}c\;=\;\bf{a^{2}\:b^{2}\;+\;b^{2}\:c^{2}\;+\;2ab^{2}c\;-\;2ab^{2}c}}$

$\\\;\rm{:\Longrightarrow\;\;(ab\;+\;bc)^{2}\;-\;2ab^{2}c\;=\;\bf{a^{2}\:b^{2}\;+\;b^{2}\:c^{2}}}$

$\\\;\underline{\boxed{\tt{(ab\;+\;bc)^{2}\;-\;2ab^{2}c\;=\;\bf{\green{a^{2}\:b^{2}\;+\;b^{2}\:c^{2}}}}}}$

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★ More Identities to know :-

$\\\;\sf{\leadsto\;\;(x\;+\;a)(x\;+\;b)\;=\;x^{2}\;+\;(a\;+\;b)x\;+\;ab}$

$\\\;\sf{\leadsto\;\;(a\;-\;b)(a\;+\;b)\;=\;a^{2}\;-\;b^{2}}$

$\\\;\sf{\leadsto\;\;(a\;+\;b\;+\;c)^{2}\;=\;a^{2}\;+\;b^{2}\;+\;c^{2}\;+\;2ab\;+\;2bc\;+\;2ac}$

$\\\;\sf{\leadsto\;\;(a\;+\;b)^{3}\;=\;a^{3}\;+\;b^{3}\;+\;3ab(a\;+\;b)}$

$\\\;\sf{\leadsto\;\;(a\;-\;b)^{3}\;=\;a^{3}\;-\;b^{3}\;-\;3ab(a\;-\;b)}$

$\\\;\sf{\leadsto\;\;(a\;-\;b\;-\;c)^{2}\;=\;a^{2}\;+\;b^{2}\;+\;c^{2}\;-\;2ab\;+\;2bc\;-\;2ac}$

$\\\;\sf{\leadsto\;\;(a\;-\;b\;+\;c)^{2}\;=\;a^{2}\;+\;b^{2}\;+\;c^{2}\;-\;2ab\;-\;2bc\;+\;2ac}$

$\\\;\sf{\leadsto\;\;a^{3}\;+\;b^{3}\;=\;(a\;+\;b)^{3}\;-\;3ab(a\;+\;b)}$

$\\\;\sf{\leadsto\;\;a^{3}\;-\;b^{3}\;=\;(a\;-\;b)^{3}\;-\;3ab(a\;-\;b)}$