Question

Expand (1/x+y/3)^3 solve it fastly

Answer 1

(1/x +y/3)³

= (1/x)³ + (y/3)³ + 3(1/X)²(y/3) + 3(1/x)(y/3)²                  
∵(a+b)³ = a³+b³+3a²b+3ab²


1/x³+y³/27 + 3(1/x²)(y/3) + 3(1/x)(y²/9)

1/x³+y³/27 + (3* y/x² * 3) + (3* y²/x * 9)    canceling 3

1/x³+y³/27 + y/x² +  y²/x * 3 

1/x³+y³/27 + y/x² +  y²/3x 

Answer 2

Given:

We have given an equation which is $(\frac{1}{x} +\frac{y}{3} )^3$.

To Find:

We have to simplify the equation by expanding it?

Step-by-step explanation:

• An cubic equation is given to us which is written below

       $(\frac{1}{x} +\frac{y}{3} )^3$

• For expanding the equation we know the formula of cube of two number which is given below

        $(a+b)^3=a^3+b^3+3a^2b+3ab^2$

• Now we will compare the given equation with the formula we will get the value of a and b which is

        $a=\frac{1}{x} ,b=\frac{y}{3}$

• Put these value in above equation and simplify it

        $(\frac{1}{x} +\frac{y}{3} )^3=(\frac{1}{x})^3 +(\frac{y}{3})^3+3(\frac{1}{x} )^2\frac{y}{3} +3\frac{1}{x} (\frac{y}{3} )^2$

• Now simplify the above expand form to get the answer

        $(\frac{1}{x} +\frac{y}{3} )^3=\frac{1}{x^3} +\frac{y^3}{27}+3(\frac{1}{x^2} )\frac{y}{3} +3\frac{1}{x} (\frac{y^2}{3} )$

• Now further simplify the terms

         $(\frac{1}{x} +\frac{y}{3} )^3=\frac{1}{x^3} +\frac{y^3}{27}+\frac{y}{x^2} +\frac{y^2}{x}$

Hence the expand form is given by the$(\frac{1}{x} +\frac{y}{3} )^3=\frac{1}{x^3} +\frac{y^3}{27}+\frac{y}{x^2} +\frac{y^2}{x}$.