Question

factorise x ki power 6 minus y ki power 6​

Answer 1

${\huge{\purple{\underline{\underline{\bf{\pink{\mathfrak{Question}}}}}}}}$

Factorise ${x}^{6}\: - \:{y}^{6}$

${\huge{\purple{\underline{\underline{\bf{\pink{\mathfrak{Solution}}}}}}}}$

⇒We know the identity

• ${a}^{3} \:−\:{b}^{3}\:=\:(\:a\:−\:b\:)(\:{a}^{2}\:+\:{b}^{2}\:+\:ab\:)$

${Using\: the \:above \:identity,\: the\: equation\: {x}^{6}−{y}^{6}}$

${\:can\: be\: factorised\: as \:follows:\:}$

⇒${x}^{6}\:−\:{y}^{6} \:$

$=\:({x}^{2})3−({y}^{2})3$

$=\:(\:{x}^{2}\:−\:y2\:)[({x}^{2})^{2}\:+\:({y}^{2})^{2}\:+\:({x}^{2}×{y}^{2})]$

$\:=\:(\:x\:−\:y\:)(\:x\:+\:y\:)(\:{x}^{2}\:+\:{y}^{2}\:+\:xy)({x}^{2}\:+\:{y}^{2}\:−\:xy\:)$

⇒$(using\:identities {a}^{2}\:−\:{b}^{2}\:=\:(\:a\:+\:b\:)(\:a\:−\:b\:) and {a}^{4}\:+\:{b}^{4}\:+\:{a}^{2}{b}^{2}\:=\:(\:{a}^{2}\:+\:{b}^{2}\:+\:ab\:)(\:{a}^{2}\:+\:{b}^{2}\:−\:ab\:) )$

$\:Hence, \:{x}^{6}\:−\:{y}^{6}\:=\:(\:x\:−\:y\:)(\:x\:+\:y\:)(\:{x}^{2}\:+\:{y}^{2}\:+\:xy)(\:{x}^{2}\:+\:{y}^{2}\:−\:xy\:)$

Answer 2

$\huge\underline\bold{AnSwEr,}$

We know the identity

${a}^{3} \:−\:{b}^{3}\:=\:(\:a\:−\:b\:)(\:{a}^{2}\:+\:{b}^{2}\:+\:ab\:)$

${Using\: the \:above \:identity,\: the\: equation\: {x}^{6}−{y}^{6}}$

${\:can\: be\: factorised\: as \:follows:\:}$

$\implies {x}^{6}\:−\:{y}^{6} \:$

$\implies \:({x}^{2})3−({y}^{2})3$

$\implies \:(\:{x}^{2}\:−\:y2\:)[({x}^{2})^{2}\:+\:({y}^{2})^{2}\:+\:({x}^{2}×{y}^{2})]$

$\implies\:(\:x\:−\:y\:)(\:x\:+\:y\:)(\:{x}^{2}\:+\:{y}^{2}\:+\:xy)({x}^{2}\:+\:{y}^{2}\:−\:xy\:)$

$(using\:identities {a}^{2}\:−\:{b}^{2}\:=\:(\:a\:+\:b\:)(\:a\:−\:b\:) \\and \:{a}^{4}\:+\:{b}^{4}\:+\:{a}^{2}{b}^{2}\:=\:(\:{a}^{2}\:+\:{b}^{2}\:+\:ab\:)(\:{a}^{2}\:+\:{b}^{2}\:−\:ab\:) )$

$\:{x}^{6}\:−\:{y}^{6}\:=\:(\:x\:−\:y\:)(\:x\:+\:y\:)(\:{x}^{2}\:+\:{y}^{2}\:+\:xy)(\:{x}^{2}\:+\:{y}^{2}\:−\:xy\:)$

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