Question

If (a+b+c)=–1 and a²+b²+c²=6, find the value of a³+b³+c³–3abc

Answer 1

${\huge{\red{\sf{Given}}}}\begin{cases}\leadsto \bf{(a+b+c)=-1}\\\leadsto \bf{a^{2}+b^{2}+c^{2}=6}\end{cases}$

${\huge{\red{\sf{To\:Find}}}}\begin{cases}\leadsto \bf{a^{3}+b^{3}+c^{3}}\end{cases}$

$\huge\red{\underline{\bf{\green{Answer}}}}$

We have,

• (a+b+c) =-1.
• $\bold{a^{2}+b^{2}+c^{2}=6}$

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Taking $\sf{\red{\mapsto (a+b+c) =-1}}$

$\sf{\green{\leadsto Squaring\:both\:sides}}$

$\sf{\implies (a+b+c) ^{2}=(-1) ^{2}}$

$\sf{\implies a^{2}+b^{2}+c^{2}+2(ab+bc+ca) =1}$

$\sf{\implies 6+2(ab+bc+ca) =1}$

$\sf{\implies 2(ab+bc+ca) =1-6}$

$\sf{\implies 2(ab+bc+ca) =5}$

$\sf{\underline{\purple{\mapsto (ab+bc+ca) =\frac{5}{2}}}}$

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Now, we know,

$\orange{\boxed{\green{\bf{a^{3}+b^{3}+c^{3}-3abc=(a+b+c) (a^{2}+b^{2}+c^{2}-ab-bc-ca) }}}}$

$\sf{\pink{\leftrightarrow Putting\:(a+b+c) =-1}}$

$\sf{\implies a^{3}+b^{3}+c^{3}-3abc=(-1) (a^{2}+b^{2}+c^{2}-ab-bc-ca)}$

$\sf{\pink{\leftrightarrow Putting\:(a^{2}+b^{2}+c^{2}) =6}}$

$\sf{a^{3}+b^{3}+c^{3}-3abc=-1(6-(ab+bc+ca) }$

$\sf{\pink{\leftrightarrow Putting\:(ab+bc+ca) =\frac{5}{2}=2.5}}$

$\sf{\implies a^{3}+b^{3}+c^{3}-3abc=-1(6-2.5) }$

$\sf{\implies a^{3}+b^{3}+c^{3}-3abc=-1\times 3.5}$

${\underline{\bf{\orange{\mapsto a^{3}+b^{3}+c^{3}-3abc=-3.5}}}}$