Question

Simplify ${{\left( 8{{p}^{-3}} \right)}^{\dfrac{2}{3}}}\times {{\left( 4{{p}^{2}} \right)}^{\dfrac{3}{2}}}\div {{p}^{-3}}$ and find the power on $p$.\[\]

Answer 1

Hint:We use the BODMAS rule to solve the expression with multiple arithmetic operations and by simplifying first bracket and then order( or power or exponent) , division, multiplication, addition, subtraction in sequence. We use the formulas of exponentiation like ${{\left( ab \right)}^{m}}={{a}^{m}}\times {{b}^{m}}$, ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$,${{a}^{m}}\div {{a}^{n}}={{a}^{m-n}}$,${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ in steps to proceed. \[\]

Complete step by step answer:
The given algebraic expression is
\[{{\left( 8{{p}^{-3}} \right)}^{\dfrac{2}{3}}}\times {{\left( 4{{p}^{2}} \right)}^{\dfrac{3}{2}}}\div {{p}^{-3}}\]
We know from the BODMAS rule that when we are given an expression with multiple arithmetic operations and then we have first simplify the terms with bracket and then order( or power or exponent) , division, multiplication, addition, subtraction in sequence. \[\]
We see that in the expression there are four operations bracket, order, multiplication, and division. So we follow the BODMAS rule and first simplify the brackets in the given expression.
We know the formula ${{\left( ab \right)}^{m}}={{a}^{m}}\times {{b}^{m}}$ for some real numbers $a,m,n,b$. We use the formula in the given expression and proceed.
\[\begin{align}
  & {{\left( 8{{p}^{-3}} \right)}^{\dfrac{2}{3}}}\times {{\left( 4{{p}^{2}} \right)}^{\dfrac{3}{2}}}\div {{p}^{-3}} \\
 & ={{8}^{\dfrac{2}{3}}}\times {{\left( {{p}^{-3}} \right)}^{\dfrac{2}{3}}}\times {{4}^{\dfrac{3}{2}}}\times {{\left( {{p}^{2}} \right)}^{\dfrac{3}{2}}}\div {{p}^{-3}} \\
\end{align}\]
We follow the BODMAS rule and now simplify the order or powers. We replace 8 with $8=2\times 2\times 2={{2}^{3}}$ and 4 with $4=2\times 2={{2}^{2}}$ in the above step.We get,
\[={{\left( {{2}^{3}} \right)}^{\dfrac{2}{3}}}\times {{\left( {{p}^{-3}} \right)}^{\dfrac{2}{3}}}\times {{\left( {{2}^{2}} \right)}^{\dfrac{3}{2}}}\times {{\left( {{p}^{2}} \right)}^{\dfrac{3}{2}}}\div {{p}^{-3}}\]
We know the formula ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ where $a,m$ or ${{a}^{m}},n$ simultaneously cannot be zero. We use the formula in the above step and get,
\[\begin{align}
  & =\left( {{2}^{3\times \dfrac{2}{3}}}\times {{p}^{-3\times \dfrac{2}{3}}} \right)\times \left( {{2}^{2\times \dfrac{3}{2}}}\times {{p}^{2\times \dfrac{3}{3}}} \right)\div {{p}^{-3}} \\
 & ={{2}^{2}}\times {{p}^{-2}}\times {{2}^{3}}\times {{p}^{3}}\div {{p}^{-3}} \\
\end{align}\]
We see that there are now two operations multiplication and division. We follow the BODMAS rule and first do the division. We know the formula for the same base $a$${{a}^{m}}\div {{a}^{n}}={{a}^{m-n}}$. We use it in the above step and get,
\[\begin{align}
  & ={{2}^{2}}\times {{p}^{-2}}\times {{2}^{3}}\times {{p}^{3-\left( -3 \right)}} \\
 & =4\times {{p}^{-2}}\times 8\times {{p}^{6}} \\
\end{align}\]
We use the commutative property of multiplication and write the numbers and the terms with power of $p$ close to each other. We have,
\[\begin{align}
  & =4\times 8\times {{p}^{-2}}\times {{p}^{6}} \\
 & =32\times {{p}^{-2}}\times {{p}^{6}} \\
\end{align}\]
We know that for the base $a$ , ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ . We use it in the above step and get,
\[\begin{align}
  & =32\times {{p}^{-2+6}} \\
 & =32{{p}^{4}} \\
\end{align}\]
So the power $p$ is 4. \[\]

Note:
We note that when we are dividing ${{p}^{-3}}$ the value of $p$ cannot be zero. We also note that in ${{a}^{m}}$, $a$ and $m$ simultaneously cannot be zero. The reciprocal or the multiplicative inverse of ${{a}^{m}}$ is given by ${{a}^{-m}}=\dfrac{1}{{{a}^{m}}}$.