Chapter 5.7, Problem 24PPSE

Solution

To calculate: The volume of the cube if the side length of the cube is $\left(x^2-\frac{1}{2}\right)$ . Rewrite the binomial as sum, consider ${\left(a+b\right)}^n$ . Identify a and b in the given binomial and which row of pascal’s triangle can be used to expand the binomial.

The volume of the cube is $x^6-\frac{3}{2}{x}^4+\frac{3}{4}{x}^2-\frac{1}{8}$ . Rewrite the binomial as sum: ${\left(x^2+\left(-\frac{1}{2}\right)\right)}^3$ , By comparing $a=x^2\text{ and }b=-\frac{1}{2}$ and use the third-row of pascal’s triangle.

Given information:

The side length of the cube is $\left(x^2-\frac{1}{2}\right)$ .

Formula used:

The volume of the cube is $V={\left(side\right)}^3$

Explanation:

Consider the given side.

Put $side=\left(x^2-\frac{1}{2}\right)$ into the volume of cube formula.

  $V={\left(x^2-\frac{1}{2}\right)}^3$

Rewrite the binomial as sum: ${\left(x^2+\left(-\frac{1}{2}\right)\right)}^3$

Compare the above binomial with ${\left(a+b\right)}^n$ .

By comparing $a=x^2\text{ and }b=-\frac{1}{2}$

The power of the given binomial is 3 so use the third-row of pascal’s triangle.

  

So, the element of the 3rd row of the pascal’s triangle are 1, 3, 3, 1. Which is the coefficient of the consecutive terms.

Binomial theorem states that:

  ${\left(a+b\right)}^n=P_0{a}^n+P_1{a}^{n-1}b+\mathrm{...}+P_n{b}^n$ here $P_0,P_1,P_2,P_3,\mathrm{...},P_n$ are the numbers in the nth row of pascal’s triangle.

Replace the coefficient of the consecutive terms.

  $\begin{array}{c}{\left(x^2-\frac{1}{2}\right)}^3={\left(x^2+\left(-\frac{1}{2}\right)\right)}^3\\ =1\cdot {\left(x^2\right)}^3{\left(-\frac{1}{2}\right)}^0+3\cdot {\left(x^2\right)}^2{\left(-\frac{1}{2}\right)}^1+3\cdot \left(x^2\right){\left(-\frac{1}{2}\right)}^2+1\cdot {\left(x^2\right)}^0{\left(-\frac{1}{2}\right)}^3\\ =x^6-\frac{3}{2}{x}^4+\frac{3}{4}{x}^2-\frac{1}{8}\end{array}$