If a=25, b=15, c=-10; then the value of frac{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We know the algebraic identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.This can also be written as: $a^{3}+b^{3}+c^{3}-3ab

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(D) We know the algebraic identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.This can also be written as: $a^{3}+b^{3}+c^{3}-3abc = \frac{1}{2}(a+b+c)[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]$.Substituting this into the given expression:$\frac{\frac{1}{2}(a+b+c)[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]}{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}} = \frac{1}{2}(a+b+c)$.Given $a=25, b=15, c=-10$,we have:$\frac{1}{2}(25+15-10) = \frac{1}{2}(30) = 15$.

If $a=25, b=15, c=-10$; then the value of $\frac{a^{3}+b^{3}+c^{3}-3abc}{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}$ is

Similar Questions

$16.02 \times 0.001 = ?$

Write down the following quotients in descending order:
$1.$ $\frac{368.39}{17}$
$2.$ $\frac{170.50}{62}$
$3.$ $\frac{875.65}{83}$

The value of $\frac{2 \frac{1}{3}-1 \frac{2}{11}}{3+\frac{1}{3+\frac{1}{3+\frac{1}{3}}}}$ is

What least value should be added to $2505$ so that it becomes a perfect square?

In a certain college,the number of girls is twice the number of boys. $\frac{1}{5}$ of the girls and $\frac{1}{8}$ of the boys took part in a social camp. What part of the total number of students took part in the camp?

Vedclass Test Series

Exam Paper Generator

Online Exam Module