Chapter 1.3, Problem 51E

To determine

To calculate: The simplified value of the expression $y^{\frac{1}{3}}\left(y^{\frac{2}{3}}+y^{\frac{5}{3}}\right)$ .

Answer to Problem 51E

The value of the expression $y^{\frac{1}{3}}\left(y^{\frac{2}{3}}+y^{\frac{5}{3}}\right)$ is $y^2+y$ .

Explanation of Solution

Given information:

The expression $y^{\frac{1}{3}}\left(y^{\frac{2}{3}}+y^{\frac{5}{3}}\right)$ .

Formula used:

To multiply two polynomial expressions, first multiply each term of first polynomial by second polynomial and then add the results. This property is known as distributive property, which is mathematically expressed as,

  $a\left(c+d\right)=ac+ad$

Law of exponent which states that to find the product of two terms with same base and different exponents, then add the exponents only keeping the base same, which can be written mathematically as,

  $x^m\cdot x^n=x^{m+n}$

Calculation:

Consider the given expression $y^{\frac{1}{3}}\left(y^{\frac{2}{3}}+y^{\frac{5}{3}}\right)$ .

Recall that to multiply two polynomial expressions, first multiply each term of first polynomial by second polynomial and then add the results. This property is known as distributive property, which is mathematically expressed as,

  $a\left(c+d\right)=ac+ad$

Apply it,

  $y^{\frac{1}{3}}\left(y^{\frac{2}{3}}+y^{\frac{5}{3}}\right)=y^{\frac{1}{3}}\cdot y^{\frac{2}{3}}+y^{\frac{1}{3}}\cdot y^{\frac{5}{3}}$

Recall the law of exponent which states that to find the product of two terms with same base and different exponents, then add the exponents only keeping the base same, which can be written mathematically as,

  $x^m\cdot x^n=x^{m+n}$

Apply it,

  $\begin{array}{c}{y}^{\frac{1}{3}}\left(y^{\frac{2}{3}}+y^{\frac{5}{3}}\right)=y^{\frac{1}{3}}\cdot y^{\frac{2}{3}}+y^{\frac{1}{3}}\cdot y^{\frac{5}{3}}\\ =y^{\frac{1}{3}+\frac{2}{3}}+y^{\frac{1}{3}+}{}^{\frac{5}{3}}\\ =y^{\frac{3}{3}}+y^{\frac{6}{3}}\\ =y+y^2\end{array}$

Thus, the value of the expression $y^{\frac{1}{3}}\left(y^{\frac{2}{3}}+y^{\frac{5}{3}}\right)$ is $y^2+y$ .