Question

Simplify using laws of exponent ${{64}^{\dfrac{-1}{3}}}\left[ {{64}^{\dfrac{1}{3}}}-{{64}^{\dfrac{2}{3}}} \right]$ \[\]

Answer 1

Hint: We follow the BODMAS rule and open the bracket using distributive law of multiplication. We then simplify the order or power on 64 using the law of product with same base ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$, the law of zero power ${{a}^{0}}=1$and the law of power of power ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ until we get numerical subtraction in integers.

Complete step-by-step answer:
We know from exponentiation that when any real number $b$ is multiplied with itself say $n$ times we can write as
\[b\times b\times ...\left( n\text{ times} \right)={{b}^{n}}\]
We read ${{b}^{n}}$ as “$b$ to the power $n.$” Here $b$ is called base and $n$ is called exponents, index, order or power. \[\]
We know from the law of product with the same base that while multiplying exponential terms with the same base we add the exponents. In symbols for some real numbers $a,m,n$
\[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\]
We know from the law of power of a power that when we raise a base with a power to another power we keep the base the same and multiply the powers. In symbols,
\[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\]
We know from the law of zero power that any number raised to the power zero is equal to 1.
\[{{a}^{0}}=1\]
We know from the BODMAS rule that when a numerical expression is given we have to first solve the bracket, then order (or power), division, multiplication, addition, subtraction. The given numerical expression is
\[{{64}^{\dfrac{-1}{3}}}\left[ {{64}^{\dfrac{1}{3}}}-{{64}^{\dfrac{2}{3}}} \right]\]
We follow the BODMAS rule and open the bracket using distributive law of addition and multiplication to have
\[={{64}^{\dfrac{-1}{3}}}\times {{64}^{\dfrac{1}{3}}}-{{64}^{\dfrac{-1}{3}}}\times {{64}^{\dfrac{2}{3}}}\]
We use the law of product with same base ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ for $b=64,m=\dfrac{-1}{3},n=\dfrac{1}{3}$ in the first term and for $b=64,m=\dfrac{-1}{3},n=\dfrac{2}{3}$ in the second term in the above step. We have
\[\begin{align}
  & ={{64}^{\dfrac{-1}{3}}}\times {{64}^{\dfrac{1}{3}}}-{{64}^{\dfrac{-1}{3}}}\times {{64}^{\dfrac{2}{3}}} \\
 & ={{64}^{\dfrac{-1}{3}+\dfrac{1}{3}}}-{{64}^{\dfrac{-1}{3}+\dfrac{2}{3}}} \\
 & ={{64}^{\dfrac{-1+1}{3}}}-{{64}^{\dfrac{-1+2}{3}}} \\
 & ={{64}^{\dfrac{0}{3}}}-{{64}^{\dfrac{1}{3}}} \\
 & ={{64}^{0}}-{{64}^{\dfrac{1}{3}}} \\
\end{align}\]
We use the law of zero power for the first term of the above step and have ${{64}^{\circ }}=1$. We proceed,
\[=1-{{64}^{\dfrac{1}{3}}}\]
We find the prime factorization of 64 and replace 64 as $64=2\times 2\times 2\times 2\times 2\times 2\times 2=4\times 4\times 4={{4}^{3}}$ in the above step. We have
\[=1-{{\left( {{4}^{3}} \right)}^{\dfrac{1}{3}}}\]
 We use the law of power of power ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ for $a=4,m=3,n=\dfrac{1}{3}$ in the above step and have,
\[=1-{{4}^{3\times \dfrac{1}{3}}}=1-{{4}^{1}}=1-4=-3\]

Note: We note that when we say ${{a}^{m}}$ , here $a$ and $m$ cannot be zero at the same time. The law of quotient with same base is given by $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ and the law of negative power $\dfrac{1}{{{a}^{m}}}={{a}^{-m}}$ where ${{a}^{-m}}$ is also called the reciprocal of ${{a}^{m}}.$ The inverse operation of exponent is logarithm.