Chapter 9.5, Problem 62E

To determine

To expand: The expression ${\left(2\sqrt{t}-1\right)}^3$ by using the Binomial Theorem.

Answer to Problem 62E

The expansion of the expression ${\left(2\sqrt{t}-1\right)}^3$ is $8t^{\frac{3}{2}}-3\left(4\right)t+6t^{\frac{1}{2}}-1$ .

Explanation of Solution

Given information: The expression is ${\left(2\sqrt{t}-1\right)}^3$ .

Theorem used: The Binomial Theorem: ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\cdots {+}_n{C}_r{x}^{n-r}{y}^r+\cdots +nx{y}^{n-1}+y^n$ , the coefficient of $x^{n-r}{y}^r$ is ${}_n{C}_r=\frac{n!}{\left(n-r\right)!r!}$ .

Calculation:

Here, $n=3$ .

First write out the Pascal’s triangle such that the row that begins with 1, 3.

  $\begin{array}{ccccccc}& & & 1& & & \\ & & 1& & 1& & \\ & 1& & 2& & 1& \\ 1& & 3& & 3& & 1\end{array}$

From the Pascal’s triangle, it is observed that the binomial coefficients of the expression are 1, 3, 3 and 1.

Expand the expression ${\left(2\sqrt{t}-1\right)}^3$ by using formula in Binomial theorem as follows.

  $\begin{array}{c}{\left(2\sqrt{t}-1\right)}^3=\left(1\right){\left(2\sqrt{t}\right)}^3+3{\left(2\sqrt{t}\right)}^{3-1}\left(-1\right)+3{\left(2\sqrt{t}\right)}^{3-2}{\left(-1\right)}^2+1{\left(-1\right)}^3\\ =8t^{\frac{3}{2}}-3\left(4\right)t+6t^{\frac{1}{2}}-1\\ =8t^{\frac{3}{2}}-12t+6t^{\frac{1}{2}}-1\end{array}$

Thus, the expansion of the expression ${\left(2\sqrt{t}-1\right)}^3$ is $8t^{\frac{3}{2}}-3\left(4\right)t+6t^{\frac{1}{2}}-1$ .