Chapter EP, Problem 11.6.10EP

To determine

Expansion of binomial ${\left(4x-a\right)}^4$ .

Answer to Problem 11.6.10EP

Expansion is $256x^4-256x^{3}a+96x^2{a}^2-16x{a}^3+a^4$

Explanation of Solution

Given information:

According to Binomial theorem if n is a natural number then the expansion of ${\left(a+b\right)}^n$ is

  ${\left(a+b\right)}^n{=}_n^{}{C}_0^{}{a}^n{b}^0{+}_n^{}{C}_1^{}{a}^{n-1}{b}^1{+}_n^{}{C}_2^{}{a}^{n-2}{b}^2\mathrm{...................}{+}_n^{}{C}_n^{}{a}^0{b}^n$

Also note ${}_n^{}{C}_0^{}=1$ and ${}_n^{}{C}_n^{}=1$ .

Need to find expansion of ${\left(4x-a\right)}^4$

Explanation:

  ${\left(4x-a\right)}^4{=}_4^{}{C}_0^{}{\left(4x\right)}^4\cdot {\left(-a\right)}^0{+}_4^{}{C}_1^{}{\left(4x\right)}^{4-1}\cdot {\left(-a\right)}^1{+}_4^{}{C}_2^{}{\left(4x\right)}^{4-2}\cdot {\left(-a\right)}^2{+}_4^{}{C}_3^{}{\left(4x\right)}^{4-3}\cdot {\left(-a\right)}^3{+}_4^{}{C}_4^{}{\left(4x\right)}^{4-4}\cdot {\left(-a\right)}^4$

  $=4^4{x}^4-\frac{4!}{\left(4-3\right)!}\cdot 64x^{3}a+\frac{4!}{2!2!}\cdot 16x^2{a}^2-\frac{4!}{\left(4-3\right)!}\cdot 4x{a}^3+a^4$

  $=256x^4-\frac{4\cdot 3!}{3!}\cdot 64x^{3}a+\frac{4\cdot 3\cdot 2!}{2!\cdot 2}\cdot 16x^2{a}^2-\frac{4!}{3!\left(4-3\right)!}\cdot 4x{a}^3+a^4$

  $=256x^4-256x^{3}a+96x^2{a}^2-16x{a}^3+a^4$

so expansion of ${\left(4x-9\right)}^4$ is .

  $=256x^4-256x^{3}a+96x^2{a}^2-16x{a}^3+a^4$