Question

if a^2 + b^2 + c^2 = 280 and ab + bc + ca = 9/2, then find the value of (a+b+c)^3

Answer 1

I hope this is correct.if it is wrong I am sorry.

Answer 2

Answer:

$(a+b+c)^3=4913$

Step-by-step explanation:

Given : If $a^2+b^2 + c^2= 280$ and $ab+bc+ca=\frac{9}{2}$

To find : The value of $(a+b+c)^3$

Solution :

We have given,

$a^2+b^2 + c^2= 280$

Expand using identity, $a^2+b^2 + c^2=(a+b+c)^2-2(ab+bc+ca)$

$(a+b+c)^2-2(ab+bc+ca)= 280$

$(a+b+c)^2-2(\frac{9}{2})= 280$

$(a+b+c)^2-9= 280$

$(a+b+c)^2= 280+9$

$(a+b+c)^2= 289$

Taking square root both side,

$(a+b+c)= \sqrt{289}$

$(a+b+c)=17$

Taking cube both side,

$(a+b+c)^3=(17)^3$

$(a+b+c)^3=4913$

Therefore, $(a+b+c)^3=4913$