Question

{a³+b³+c³-3abc}÷{a+b+c}​

Answer 1

$\underline { \blue { By \: Algebraic \:Identity}}$

$\boxed { \pink { x^{3} + y^{3} + z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)}}$

$Now,\red{\frac{a^{3}+b^{3}+c^{3}-3abc}{(a+b+c)} }$

$= \frac{(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca}{(a+b+c)} \\= (a^{2}+b^{2}+c^{2}-ab-bc-ca)$

Therefore.,

$\red{\frac{a^{3}+b^{3}+c^{3}-3abc}{(a+b+c)} }$

$\green {= (a^{2}+b^{2}+c^{2}-ab-bc-ca)}$

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