Question

Question: Let x and y be real numbers. Find the coefficient of${\mathit{x}}^{\mathbf{12}}{\mathit{y}}^{\mathbf{6}}$ in the expansion of ${\left(x^3-3y\right)}^{\mathbf{10}}$.

Short answer

Answer:

The coefficient of $x^{12}{y}^6$in the expansion of ${\left(x^3-3y\right)}^{10}$is1,53,090.

Step-by-step solution

Definition of the Binomial Theorem

This theorem states that let R be a commutative rind and$\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{R}$ . Then each positive integer n,

${\left(a+b\right)}^{\mathbf{n}}\mathbf{=}{\mathit{a}}^{\mathbf{n}}\mathbf{+}\left[\begin{array}{c}n\\ 1\end{array}\right]{\mathit{a}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathit{b}\mathbf{+}\left[\begin{array}{c}n\\ 1\end{array}\right]{\mathit{a}}^{\mathbf{n}\mathbf{-}\mathbf{2}}{\mathit{b}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}\left[\begin{array}{c}n\\ n-1\end{array}\right]\mathbf{}\mathit{a}{\mathit{b}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{+}{\mathit{b}}^{\mathbf{n}}$

Find the coefficient of x12y6 in the expansion of (x3-3y)10

Consider the equation as follows:

${\left(x^3-3y\right)}^{10}$

Let $a=x^3$and b = 3y.

Apply the binomial theorem and substitute the corresponding values.

Therefore, the coefficient of $x^{12}{y}^6$in the expansion of ${\left(x^3-3y\right)}^{10}$is 1,53,090.