Question

if a+b+c=15 and a 2 +b 2 +c 2 =83, then find the value of a 3 +b 3 +c 3 -3abc

Answer 1

Answer:

a³+b³+c³-3abc = 180

Explanation:

Given,

a+b+c = 15 ----(1)

a²+b²+c² = 83 ---(2)

We know the algebraic identity:

a²+b²+c²+2(ab+bc+ca) = (a+b+c)²

=> 83 + 2(ab+bc+ca) = 15²

/* from (1) & (2) */

=> 2(ab+bc+ca) = 225 - 83

=> 2(ab+bc+ca) = 142

=> ab+bc+ca = 142/2

=> ab+bc+ca = 71 ----(3)

Now ,

By the algebraic identity :

a³+b³+c³-3abc

= (a+b+c)[a²+b²+c²-(ab+bc+ca)]

= 15(83-71)

/*From (1),(2)&(3) */

= 15 × 12

= 180

Therefore,

a³+b³+c³-3abc = 180

••••

Answer 2

180 is the value of $\bold{a^{3}+b^{3}+c^{3}-3 a b c}$

Given:

$a+b+c=15 \\a^2  +b^2  +c^2  =83$

To find:

The value of $a^{3}+b^{3}+c^{3}-3 a b c =?$

Solution:

Given, $a+b+c=15 and a^{2}+b^{2}+c^{2}=83$

$a+b+c=15$

Squaring both sides of the above equation, we get

$\begin{array}{l}{(a+b+c)^{2}=15^{2}} \\ {a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=225} \\ {83+2(a b+b c+c a)=225} \\ {2(a b+b c+c a)=225-83}\end{array}$

$\begin{array}{l}{2(a b+b c+c a)=142} \\ {a b+b c+c a=\frac{142}{2}} \\ {a b+b c+c a=71}\end{array}$

Now, we know,

$\begin{aligned} a^{3}+b^{3}+c^{3}-3 a b c &=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) \\ &=(a+b+c)\left(a^{2}+b^{2}+c^{2}-(a b+b c+c a)\right) \end{aligned}$

Putting the values of $a+b+c  and  a^{2}+b^{2}+c^{2}  and  ab-bc-ca,$ we get,

$\bold{\begin{array}{l}{=(15)(83-71)} \\ {=15 \times 12} \\ {=180}\end{array}}$

Therefore, the value of $\bold{a^{3}+b^{3}+c^{3}-3 a b c is 180.}$