Question

(3a/3b)a+b(3b/3c)b+c(3c/3a)c+a=1​

Answer 1

Given:

$\[(3a/3b)a + b\left( {3b/3c} \right)b + c\left( {3c/3a} \right)c + a = 1\]$

To Find:

Simplify above given equation.

Solution:

$\[(3a/3b)a + b\left( {3b/3c} \right)b + c\left( {3c/3a} \right)c + a = 1\]$

On simplifying;

$\dfrac{a^2}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}+a=1\\\dfrac{a^3c+b^4a+c^4b+a^2bc}{abc}=1\\\dfrac{(a^3c+a^2bc)+(b^4a+c^4b)}{abc}=1\\\dfrac{a^2c(a+b)+b(b^3a+c^4)}{abc}=1\\a^2c(a+b)+b(b^3a+c^4)=abc\\a^2c(a+b)+b(b^3a+c^4)-abc=0$

Hence, $a^2c(a+b)+b(b^3a+c^4)-abc=0$ is the simplified equation of given above equation.

Answer 2

Answer:

3^0

because

all the power will be simplified