Question

Prove that the following equation is correct \[\dfrac{{{3^{ - 3}} \times {6^2} \times \sqrt {98} }}{{{5^2} \times \sqrt[3]{{\dfrac{1}{{25}}}} \times {{\left( {15} \right)}^{ - \dfrac{4}{3}}} \times {3^{\dfrac{1}{3}}}}} = 28\sqrt 2 \]

Answer 1

Hint: You cannot do anything with two terms having different power and base. First, start with changing bases into prime numbers using property like ${a^m} \times {b^m} = {\left( {ab} \right)^m}$. Simplify fractional base and radical signs using ${a^{ - b}} = \dfrac{1}{{{a^b}}}$ and $\sqrt[b]{a} = {a^{\dfrac{1}{b}}}$ . Now combine the powers of the same base using ${a^m} \times {a^n} = {a^{m + n}}$ . Simplify it further by taking all terms to the numerator to prove the required answer.

Complete step-by-step answer:
In this problem, we need to prove that the expression in the left-hand side (LHS) of the equation is equal to the right-hand side (RHS). The expression on the left-side is a fraction that can be solved using the properties of surds and indices.
Before starting with the solution, let’s first understand a few properties that will be useful for this problem.
Properties:
${a^{ - b}} = \dfrac{1}{{{a^b}}}$.........………. (i)
$\sqrt[b]{a} = {a^{\dfrac{1}{b}}}$...........……….. (ii)
${a^m} \times {b^m} = {\left( {ab} \right)^m}$............………… (iii)
${a^m} \times {a^n} = {a^{m + n}}$...........……….. (iv)
${\left( {{a^m}} \right)^n} = {a^{m \times n}}$...........………… (v)
Now, let’s consider LHS and try to separate the prime factors in both numerator and denominator
$ \Rightarrow \dfrac{{{3^{ - 3}} \times {6^2} \times \sqrt {98} }}{{{5^2} \times \sqrt[3]{{\dfrac{1}{{25}}}} \times {{\left( {15} \right)}^{ - \dfrac{4}{3}}} \times {3^{\dfrac{1}{3}}}}} = \dfrac{{{3^{ - 3}} \times {{\left( {3 \times 2} \right)}^2} \times \sqrt {\left( {2 \times {7^2}} \right)} }}{{{5^2} \times \sqrt[3]{{\dfrac{1}{{\left( {{5^2}} \right)}}}} \times {{\left( {3 \times 5} \right)}^{ - \dfrac{4}{3}}} \times {3^{\dfrac{1}{3}}}}}$
So, now we have an expression that has power in the prime numbers which are $2{\text{, }}3{\text{, }}5{\text{ and }}7$ . Let’s use the properties (i), (ii), (iii), (iv) and (v) to separate terms in numerator and denominator.
$ \Rightarrow \dfrac{{{3^{ - 3}} \times {{\left( {3 \times 2} \right)}^2} \times \sqrt {\left( {2 \times {7^2}} \right)} }}{{{5^2} \times \sqrt[3]{{\dfrac{1}{{\left( {{5^2}} \right)}}}} \times {{\left( {3 \times 5} \right)}^{ - \dfrac{4}{3}}} \times {3^{\dfrac{1}{3}}}}} = \dfrac{{{3^{ - 3}} \times {3^2} \times {2^2} \times {2^{\dfrac{1}{2}}} \times {{\left( {{7^2}} \right)}^{\dfrac{1}{2}}}}}{{{5^2} \times {5^{ - \dfrac{2}{3}}} \times {3^{ - \dfrac{4}{3}}} \times {5^{ - \dfrac{4}{3}}} \times {3^{\dfrac{1}{3}}}}}$
Now, for using the property (iv), we combine the powers with the same base together. This gives us:
\[ \Rightarrow \dfrac{{{3^{ - 3}} \times {3^2} \times {2^2} \times {2^{\dfrac{1}{2}}} \times {{\left( {{7^2}} \right)}^{\dfrac{1}{2}}}}}{{{5^2} \times {5^{ - \dfrac{2}{3}}} \times {3^{ - \dfrac{4}{3}}} \times {5^{ - \dfrac{4}{3}}} \times {3^{\dfrac{1}{3}}}}} = \dfrac{{\left( {{3^{ - 3}} \times {3^2}} \right) \times \left( {{2^2} \times {2^{\dfrac{1}{2}}}} \right)}}{{\left( {{5^2} \times {5^{ - \dfrac{2}{3}}} \times {5^{ - \dfrac{4}{3}}}} \right) \times \left( {{3^{ - \dfrac{4}{3}}} \times {3^{\dfrac{1}{3}}}} \right)}}\]
So, by solving each of the parentheses and using relation (iv) in all of these, we get:
\[ \Rightarrow \dfrac{{\left( {{3^{ - 3}} \times {3^2}} \right) \times \left( {{2^2} \times {2^{\dfrac{1}{2}}}} \right) \times 7}}{{\left( {{5^2} \times {5^{ - \dfrac{2}{3}}} \times {5^{ - \dfrac{4}{3}}}} \right) \times \left( {{3^{ - \dfrac{4}{3}}} \times {3^{\dfrac{1}{3}}}} \right)}} = \dfrac{{\left( {{3^{ - 3 + 2}}} \right) \times \left( {{2^{2 + \dfrac{1}{2}}}} \right) \times 7}}{{\left( {{5^{2 - \dfrac{2}{3} - \dfrac{4}{3}}}} \right) \times \left( {{3^{ - \dfrac{4}{3} + \dfrac{1}{3}}}} \right)}} = \dfrac{{\left( {{3^{ - 1}}} \right) \times \left( {{2^{\dfrac{5}{2}}}} \right) \times 7}}{{\left( {{5^0}} \right) \times \left( {{3^{ - 1}}} \right)}}\]
Also, we know that the power zero to any number will give us one, i.e. ${a^0} = 1$ . Try to transpose ${3^{ - 1}}$ to the numerator and then combine indices of \[3\] using property (iv) to get:
$ \Rightarrow \dfrac{{\left( {{3^{ - 1}}} \right) \times \left( {{2^{\dfrac{5}{2}}}} \right) \times 7}}{{\left( {{5^0}} \right) \times \left( {{3^{ - 1}}} \right)}} = \dfrac{{{3^{ - 1 + 1}} \times {2^{\dfrac{5}{2}}} \times 7}}{1} = {3^0} \times {2^{\dfrac{5}{2}}} \times 7$
Now we can put ${3^0} = 1$ and also we can distribute the index using the property (iv) ${2^{\dfrac{5}{2}}} = {2^{2 + \dfrac{1}{2}}} = {2^2} \times {2^{\dfrac{1}{2}}}$
Therefore, after this we will get:
$ \Rightarrow {3^0} \times {2^{\dfrac{5}{2}}} \times 7 = 1 \times {2^2} \times {2^{\dfrac{1}{2}}} \times 7 = 4 \times 7 \times {2^{\dfrac{1}{2}}}$
Now we can use property (ii) to change fractional power into the radical sign, so we get:
$ \Rightarrow 4 \times 7 \times {2^{\dfrac{1}{2}}} = 4 \times 7 \times \sqrt 2 = 28\sqrt 2 $
Hence, we proved that the given equation \[\dfrac{{{3^{ - 3}} \times {6^2} \times \sqrt {98} }}{{{5^2} \times \sqrt[3]{{\dfrac{1}{{25}}}} \times {{\left( {15} \right)}^{ - \dfrac{4}{3}}} \times {3^{\dfrac{1}{3}}}}} = 28\sqrt 2 \] is correct.

Note: The use of the properties of indices and surds played an important role in the solution. In problems like this, you should always try to consider one side of the equation and figure out ways to transform it into an expression the same as right-side. Remember that we cannot simplify two terms with different surds and different indices. So to solve such an expression, focus to change it into similar indices or similar surd.