Question

If a + b + c = 16 and a2 + b2 + c2 = 90, then find the value of a3 + b3 + c3 – 3abc.​

Answer 1

Answer:

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Given= a+b+c = 16

${a}^{2} + {b}^{2} + {c}^{2} = 90$

To find= Value of${a}^{3} + {b}^{3} + {c}^{3} - 3abc$

SOLUTION:- As we know,

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

$=> {(12)}^{2} = 90 + 2(ab + bc + ca)$

$=> 144 = 90 + 2(ab + bc + ca)$

$=> 2(ab + bc + ca) = 54$

$=> ab + bc + ca = 27$

Now,${a}^{3} + {b}^{3} + {c}^{3} - 3abc$

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

$= 12 \times (90 - 27)$

$= 12 \times 63$

$= 756$

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Answer 2

Given:

$a+b+c=16\\a^2+b^2+c^2=90$

To find:

$a^3+b^3+c^3-3abc=?$

Solution:

$\to a+b+c=16....(a)\\\to a^2+b^2+c^2=90.....(b)$

Formula:

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

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

whole square the equation (a) and put the given value:

$\Rightarrow (a+b+c)^2=(a^2+b^2+c^2)+2(ab+bc+ca)\\\\\Rightarrow (16)^2=(90)+2(ab+bc+ca)\\\\\Rightarrow (256-90)=2(ab+bc+ca)\\\\\Rightarrow (ab+bc+ca)= \frac{256-90}{2}\\\\\Rightarrow (ab+bc+ca)= \frac{166}{2}\\\\\Rightarrow (ab+bc+ca)= 83$

So, The value of $\bold{a^3+b^3+c^3-3abc}$ is:

$\Rightarrow (a+b+c)(a^2+b^2+c^2 -ab-bc-ca)\\\\\Rightarrow (a+b+c)(a^2+b^2+c^2 -(ab+bc+ca))\\\\\Rightarrow 16 \times (90- 83)\\\\\Rightarrow 16 \times 7 \\\\\boxed {\Rightarrow 112}$

The final value is 112.