# If a+b+c=7, a^2+b^2+c^2=35 and a^3+b^3+c^3=151, then what is abc ?

Apr 17, 2017

$\left(a b c\right) = - 15$

#### Explanation:

$a + b + c = 7$
$5 + 3 + \left(- 1\right) = 7$

${a}^{2} + {b}^{2} + {c}^{2} = 35$
${5}^{2} + {3}^{2} + {\left(- 1\right)}^{2}$
$= 25 + 9 + 1$
$= 35$

${a}^{3} + {b}^{3} + {c}^{3} = 151$
${5}^{3} + {3}^{3} + {\left(- 1\right)}^{3}$
$= 125 + 27 - 1$
$= 151$

$\left(a b c\right)$
=5×3×-1
$= - 15$

Apr 17, 2017

$a b c = - 15$

#### Explanation:

Given:

$\left\{\begin{matrix}a + b + c = 7 \\ {a}^{2} + {b}^{2} + {c}^{2} = 35 \\ {a}^{3} + {b}^{3} + {c}^{3} = 151\end{matrix}\right.$

Note that:

${\left(a + b + c\right)}^{3}$

$= {a}^{3} + {b}^{3} + {c}^{3} + 3 a {b}^{2} + 3 {a}^{2} b + 3 b {c}^{2} + 3 {b}^{2} c + 3 c {a}^{2} + 3 a {c}^{2} + 6 a b c$

$\left(a + b + c\right) \left({a}^{2} + {b}^{2} + {c}^{2}\right)$

$= {a}^{3} + {b}^{3} + {c}^{3} + a {b}^{2} + {a}^{2} b + b {c}^{2} + {b}^{2} c + c {a}^{2} + {c}^{2} a$

So:

$6 a b c = {\left(a + b + c\right)}^{3} - 3 \left(a + b + c\right) \left({a}^{2} + {b}^{2} + {c}^{2}\right) + 2 \left({a}^{3} + {b}^{3} + {c}^{3}\right)$

$\textcolor{w h i t e}{6 a b c} = {7}^{3} - 3 \left(7\right) \left(35\right) + 2 \left(151\right)$

$\textcolor{w h i t e}{6 a b c} = 343 - 735 + 302$

$\textcolor{w h i t e}{6 a b c} = - 90$

Dividing both sides by $6$ we find:

$a b c = - 15$

Apr 17, 2017

$- 15.$

#### Explanation:

We will use the following Result :

${a}^{3} + {b}^{3} + {c}^{3} - 3 a b c$

$= \left(a + b + c\right) \left({a}^{2} + {b}^{2} + {c}^{2} - a b - b c - c a\right) \ldots \ldots \ldots \ldots \ldots \left(\star\right)$

Now, $a + b + c = 7 \Rightarrow {\left(a + b + c\right)}^{2} = 49.$

$\Rightarrow {a}^{2} + {b}^{2} + {c}^{2} + 2 \left(a b + b c + c a\right) = 49.$

But, we are given that, ${a}^{2} + {b}^{2} + {c}^{2} = 35.$

$\therefore 35 + 2 \left(a b + b c + c a\right) = 49 , \mathmr{and} ,$

$a b + b c + c a = \frac{49 - 35}{2} = 7. . . \left(\ast\right)$

Thus, altogether, we have,

${a}^{3} + {b}^{3} + {c}^{3} = 151 , a + b + c = 7 , {a}^{2} + {b}^{2} + {c}^{2} = 35. . . \text{(given),}$

&, by (ast), ab+bc+ca=7.

Using all these in $\left(\star\right) ,$ we get,

$151 - 3 a b c = \left(7\right) \left\{35 - \left(7\right)\right\} = 196.$

$\therefore a b c = \frac{1}{3} \left(151 - 196\right) = - \frac{45}{3} = - 15 ,$ as deived by,

Respcted George C. Sir.

Enjoy Maths.!