Question

Find the value of $\dfrac{4}{{{{216}^{\dfrac{{ - 2}}{3}}}}} + \dfrac{1}{{{{256}^{\dfrac{{ - 1}}{2}}}}} - \dfrac{2}{{{{343}^{\dfrac{{ - 1}}{3}}}}}$

Answer 1

Hint:Here the actual problem is about the denominator part of each term. So, we will simplify the denominator part of each term and then will solve the whole problem. We will take denominators of the three terms separately, and simplify them to the simplest fraction term. Then we will substitute them in their respective place in the question. Then we will again simplify the whole question. We will do this using some formulae such as, ${a^{ - 1}} = \dfrac{1}{a}$, ${\left( {{a^m}} \right)^n} = {a^{m \times n}}$

Complete step-by-step answer:
Here they given $\dfrac{4}{{{{216}^{\dfrac{{ - 2}}{3}}}}} + \dfrac{1}{{{{256}^{\dfrac{{ - 1}}{2}}}}} - \dfrac{2}{{{{343}^{\dfrac{{ - 1}}{3}}}}}$
Solving denominators part first,
The denominator of first term $\dfrac{4}{{{{216}^{\dfrac{{ - 2}}{3}}}}}$ is ${216^{\dfrac{{ - 2}}{3}}}$
Converting it into general form we get,
${216^{\dfrac{{ - 2}}{3}}}$=${(\dfrac{1}{{216}})^{\dfrac{2}{3}}}$
To cancel the denominator of the power we can write 216 as the cube of 6.
${(\dfrac{1}{{216}})^{\dfrac{2}{3}}} = {[{(\dfrac{1}{6})^3}]^{\dfrac{2}{3}}}$
Using the formula ${\left( {{a^m}} \right)^n} = {a^{m \times n}}$
${[{(\dfrac{1}{6})^3}]^{\dfrac{2}{3}}} = {(\dfrac{1}{6})^{3 \times \dfrac{2}{3}}}$
Cancelling 3 upon 3 and converting it into general form we get,
$ = \dfrac{1}{{36}}$………….(1)
Similarly the denominator of second term $\dfrac{1}{{{{256}^{\dfrac{{ - 1}}{2}}}}}$ is ${256^{\dfrac{{ - 1}}{2}}}$
${256^{\dfrac{{ - 1}}{2}}} = {(\dfrac{1}{{256}})^{\dfrac{1}{2}}}$
To cancel the denominator of the power we can write 256 as a square of 16.
${(\dfrac{1}{{256}})^{\dfrac{1}{2}}} = {[{(\dfrac{1}{{16}})^2}]^{\dfrac{1}{2}}}$
Cancelling 2 upon 2 we get,
${(\dfrac{1}{{16}})^{2 \times }}^{\dfrac{1}{2}} = {(\dfrac{1}{{16}})^1}$
$ = \dfrac{1}{{16}}$…………………..(2)
Similarly the denominator of third term $\dfrac{2}{{{{343}^{\dfrac{{ - 1}}{3}}}}}$ is ${343^{\dfrac{{ - 1}}{3}}}$
${343^{\dfrac{{ - 1}}{3}}} = {(\dfrac{1}{{343}})^{\dfrac{1}{3}}}$
To cancel the denominator of the power we can write 343 as cube of 7
${(\dfrac{1}{{343}})^{\dfrac{1}{3}}} = {[{(\dfrac{1}{7})^3}]^{\dfrac{1}{3}}}$
Cancelling 3 upon 3 we get,
${[{(\dfrac{1}{7})^3}]^{\dfrac{1}{3}}} = {(\dfrac{1}{7})^{^1}}$
$ = \dfrac{1}{7}$……………….(3)
Putting equations (1), (2), (3) in the given question,
$\dfrac{4}{{{{216}^{\dfrac{{ - 2}}{3}}}}} + \dfrac{1}{{{{256}^{\dfrac{{ - 1}}{2}}}}} - \dfrac{2}{{{{343}^{\dfrac{{ - 1}}{3}}}}}$
$ = \dfrac{4}{{\dfrac{1}{{36}}}} + \dfrac{1}{{\dfrac{1}{6}}} - \dfrac{2}{{\dfrac{1}{7}}}$
As there are dfractions in denominator, multiplying their reciprocal terms with respective numerator we get,
$ = (4 \times 36) + (1 \times 6) - (1 \times 7)$
=144+6-7
=143
Hence the value of $\dfrac{4}{{{{216}^{\dfrac{{ - 2}}{3}}}}} + \dfrac{1}{{{{256}^{\dfrac{{ - 1}}{2}}}}} - \dfrac{2}{{{{343}^{\dfrac{{ - 1}}{3}}}}}$ is 143.

Note:Students can also simplify the denominators of each term in the given problem while doing solution.They might make a mistake while dividing with a fraction.They can also simplify all the denominators by not converting it into general form or function form.Students should remember all exponent rules and formulas for solving these types of questions.