If a+b+c=5 and ab+bc+ca=10, then the value of a3+b3+c3-3abc is

# If $a+b+c=5$ and $\mathit{ab}+\mathit{bc}+\mathit{ca}=10$, then the value of ${a}^{3}+{b}^{3}+{c}^{3}-3\mathit{abc}$ is

1. A
25
2. B
-25
3. C
35
4. D
-35

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### Solution:

It is given that $a+b+c=5$ and $\mathit{ab}+\mathit{bc}+\mathit{ca}=10$.
We know ${\left(a+b+c\right)}^{2}={a}^{2}+{b}^{2}+{c}^{2}+2\left(\mathit{ab}+\mathit{bc}+\mathit{ca}\right)$.
Substituting values, $a+b+c=5$ and $\mathit{ab}+\mathit{bc}+\mathit{ca}=10$ in the above identity,
${5}^{2}={a}^{2}+{b}^{2}+{c}^{2}+2\left(10\right)$
$25={a}^{2}+{b}^{2}+{c}^{2}+20$
$25-20={a}^{2}+{b}^{2}+{c}^{2}$
$5={a}^{2}+{b}^{2}+{c}^{2}$
$⇒{a}^{2}+{b}^{2}+{c}^{2}=5$          …….eq(i)
Now, we know ${a}^{3}+{b}^{3}+{c}^{3}-3\mathit{abc}=\left(a+b+c\right)\left({a}^{2}+{b}^{2}+{c}^{2}-\mathit{ab}-\mathit{bc}-\mathit{ca}\right).$ Substituting values, $a+b+c=5$, $\mathit{ab}+\mathit{bc}+\mathit{ca}=10$ and ${a}^{2}+{b}^{2}+{c}^{2}=5$ in above identity,
${a}^{3}+{b}^{3}+{c}^{3}-3\mathit{abc}=\left(a+b+c\right)\left({a}^{2}+{b}^{2}+{c}^{2}-\mathit{ab}-\mathit{bc}-\mathit{ca}\right)$
${a}^{3}+{b}^{3}+{c}^{3}-3\mathit{abc}=\left(5\right)\left(5-10\right)$
${a}^{3}+{b}^{3}+{c}^{3}-3\mathit{abc}=\left(5\right)\left(-5\right)$
${a}^{3}+{b}^{3}+{c}^{3}-3\mathit{abc}=-25$
Therefore, option 2 is correct.

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