frac{1}{2}(a+b+c){(a-b)^{2}+(b-c)^{2}+(c-a)^{ | vedclass

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

Question

(B) We know the standard algebraic identity:$a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$Multiplying and dividing the right side by $

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We know the standard algebraic identity:$a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$Multiplying and dividing the right side by $2$,we get:$a^{3}+b^{3}+c^{3}-3abc = \frac{1}{2}(a+b+c)(2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca)$This can be rewritten as:$a^{3}+b^{3}+c^{3}-3abc = \frac{1}{2}(a+b+c)\{(a^{2}-2ab+b^{2})+(b^{2}-2bc+c^{2})+(c^{2}-2ca+a^{2})\}$$a^{3}+b^{3}+c^{3}-3abc = \frac{1}{2}(a+b+c)\{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\}$Therefore,the given expression is equal to $a^{3}+b^{3}+c^{3}-3abc$.

$\frac{1}{2}(a+b+c)\{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\}=?$

Similar Questions

Let $\lambda \neq 0$ be in $\mathbb{R}$. If $\alpha$ and $\beta$ are the roots of the equation $x^{2}-x+2 \lambda=0$ and $\alpha$ and $\gamma$ are the roots of the equation $3x^{2}-10x+27 \lambda=0$,then $\frac{\beta \gamma}{\lambda}$ is equal to:

If $x^{2}+y^{2}+1=2x,$ then the value of $x^{3}+y^{5}$ is

Two candidates attempt to solve the equation $x^2 + px + q = 0$. One starts with a wrong value of $p$ and finds the roots to be $2$ and $6$,and the other starts with a wrong value of $q$ and finds the roots to be $2$ and $-9$. The roots of the original equation are

Construct a quadratic equation whose roots have the sum $= 6$ and product $= -16$.

The sum of the roots of a quadratic equation is $2$ and the sum of their cubes is $98$. Then the equation is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module