# Finding Binomial Expansion of $(11x+y)^3$

Utilize the Binomial Expansion Calculator and enter your input term in the input field ie., $(11x+y)^3$ & press the calculate button to get the result ie., $1331x^3 + 363x^2y + 33xy^2 + y^3$ along with a detailed solution in a fraction of seconds.

Ex: (x+1)^2 (or) (x+7)^7 (or) (x+3)^4

Binomial Expansion of:

## Elaborate Steps to Expand $(11x+y)^3$ Using Binomial Theorem

According to the binomial formula $(a+b)^n$ = $\sum_{k=0}^{n} {^nC_k}(a^{n-k}b^{k})$

So $(11.x + y)^3$ = $\sum_{k=0}^3 {^3C_k}((11.x)^{3-k}(y)^{k})$

By expanding the summation:

$\frac{3!}{(3-0)!0!}(11.x)^{3-0}\times{}(y)^0+\frac{3!}{(3-1)!1!}(11.x)^{3-1}\times{}(y)^1+\frac{3!}{(3-2)!2!}(11.x)^{3-2}\times{}(y)^2+\frac{3!}{(3-3)!3!}(11.x)^{3-3}\times{}(y)^3$

$= \frac{6}{(6)1}(11.x)^{3-0}\times{}(y)^0+\frac{6}{(2)1}(11.x)^{3-1}\times{}(y)^1+\frac{6}{(1)2}(11.x)^{3-2}\times{}(y)^2+\frac{6}{(1)6}(11.x)^{3-3}\times{}(y)^3$

$= 1(11.x)^{3-0}\times{}(y)^0+3(11.x)^{3-1}\times{}(y)^1+3(11.x)^{3-2}\times{}(y)^2+1(11.x)^{3-3}\times{}(y)^3$

$= (11.x)^{3-0}\times{}(y)^0+(3)(11.x)^{3-1}\times{}(y)^1+(3)(11.x)^{3-2}\times{}(y)^2+(11.x)^{3-3}\times{}(y)^3$

$= (11.x)^{3}\times{}(y)^0+(3)(11.x)^{2}\times{}(y)^1+(3)(11.x)^{1}\times{}(y)^2+(11.x)^{0}\times{}(y)^3$

$= (11.x)^{3}\times{}1+(3)(11.x)^{2}\times{}(y)^1+(3)(11.x)^{1}\times{}(y)^2+1\times{}(y)^3$

$= 1331.x^3\times{}(1)+(3)121.x^2\times{}(y)+(3)11.x\times{}(y^2)+1\times{}(y^3)$

$= 1331x^3 + 363x^2y + 33xy^2 + y^3$

### FAQs on Binomial Expansion of $(11x+y)^3$

1. How to simplify the Binomial Expansion $(11x+y)^3$?

You can expand the given term $(11x+y)^3$ in a binomial expansion by using Newton's binomial theorem & the formula of it.

2. What is the Binomial Expansion of $(11x+y)^3$?

The Binomial Expansion of $(11x+y)^3$ is $1331x^3 + 363x^2y + 33xy^2 + y^3$.

3. Where can I obtain a step by step solution to expand the given binomial $(11x+y)^3$?

You can obtain the step by step solution for Binomial Expansion of $(11x+y)^3$ on our page.