Question

Solve the given equation: $2\sqrt{2}{{a}^{3}}+3\sqrt{3}{{b}^{3}}+6\sqrt{3}{{a}^{2}}b+9\sqrt{2}a{{b}^{2}}$?

Answer 1

Hint: We start solving the problem by assuming that the given equation is equal to variable x. We then use $c\sqrt{d}=\sqrt{{{c}^{2}}d}$ to proceed through the problem. We then use the law of exponents ${{\left( {{a}^{m}} \right)}^{n}}={{\left( {{a}^{n}} \right)}^{m}}$, ${{a}^{m}}.{{b}^{m}}={{\left( ab \right)}^{m}}$ to proceed further through the problem. After making certain arrangements in the equation, we can see that x resembles with ${{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}={{\left( a+b \right)}^{3}}$ which leads us to the required solution.

Complete step-by-step solution:
According to the problem, we need to find solve the given equation $2\sqrt{2}{{a}^{3}}+3\sqrt{3}{{b}^{3}}+6\sqrt{3}{{a}^{2}}b+9\sqrt{2}a{{b}^{2}}$. Let us assume this is equal to x.
So, we have $x=2\sqrt{2}{{a}^{3}}+3\sqrt{3}{{b}^{3}}+6\sqrt{3}{{a}^{2}}b+9\sqrt{2}a{{b}^{2}}$ ---(1).
We know that $c\sqrt{d}=\sqrt{{{c}^{2}}d}$, we substitute this in equation (1).
$\Rightarrow x=\left( \sqrt{{{2}^{2}}\times 2} \right){{a}^{3}}+\left( \sqrt{{{3}^{2}}\times 3} \right){{b}^{3}}+6\sqrt{3}{{a}^{2}}b+9\sqrt{2}a{{b}^{2}}$.
$\Rightarrow x=\left( \sqrt{{{2}^{3}}} \right){{a}^{3}}+\left( \sqrt{{{3}^{3}}} \right){{b}^{3}}+6\sqrt{3}{{a}^{2}}b+9\sqrt{2}a{{b}^{2}}$ ---(2).
We know that ${{\left( {{a}^{m}} \right)}^{n}}={{\left( {{a}^{n}} \right)}^{m}}$, we use this in equation (2).
$\Rightarrow x={{\left( \sqrt{2} \right)}^{3}}{{a}^{3}}+{{\left( \sqrt{3} \right)}^{3}}{{b}^{3}}+6\sqrt{3}{{a}^{2}}b+9\sqrt{2}a{{b}^{2}}$ ---(3).
We know that ${{a}^{m}}.{{b}^{m}}={{\left( ab \right)}^{m}}$. We use this in equation (3).
$\Rightarrow x={{\left( \sqrt{2}a \right)}^{3}}+{{\left( \sqrt{3}b \right)}^{3}}+6\sqrt{3}{{a}^{2}}b+9\sqrt{2}a{{b}^{2}}$.
$\Rightarrow x={{\left( \sqrt{2}a \right)}^{3}}+{{\left( \sqrt{3}b \right)}^{3}}+3\left( 2{{a}^{2}} \right)\left( \sqrt{3}b \right)+3\left( \sqrt{2}a \right)\left( 3{{b}^{2}} \right)$.
$\Rightarrow x={{\left( \sqrt{2}a \right)}^{3}}+{{\left( \sqrt{3}b \right)}^{3}}+3\left( {{\left( \sqrt{2} \right)}^{2}}{{a}^{2}} \right)\left( \sqrt{3}b \right)+3\left( \sqrt{2}a \right)\left( {{\left( \sqrt{3} \right)}^{2}}{{b}^{2}} \right)$---(4).
We know that ${{a}^{m}}.{{b}^{m}}={{\left( ab \right)}^{m}}$. We use this in equation (4).
$\Rightarrow x={{\left( \sqrt{2}a \right)}^{3}}+{{\left( \sqrt{3}b \right)}^{3}}+3{{\left( \sqrt{2}a \right)}^{2}}\left( \sqrt{3}b \right)+3\left( \sqrt{2}a \right){{\left( \sqrt{3}b \right)}^{2}}$ ---(5).
Let us assume $\sqrt{2}a=d$ and $\sqrt{3}b=e$. Let us use this in equation (5).
So, we have $x={{d}^{3}}+{{e}^{3}}+3{{d}^{2}}e+3d{{e}^{2}}$ ---(6).
We can see that $x$ resembles with ${{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}={{\left( a+b \right)}^{3}}$. We substitute this in equation (6).
$\Rightarrow x={{\left( d+e \right)}^{3}}$ ---(7).
Now, we replace the values of d and e in equation (7).
$\Rightarrow x={{\left( \sqrt{2}a+\sqrt{3}b \right)}^{3}}$.
We have solved the equation $2\sqrt{2}{{a}^{3}}+3\sqrt{3}{{b}^{3}}+6\sqrt{3}{{a}^{2}}b+9\sqrt{2}a{{b}^{2}}$ as ${{\left( \sqrt{2}a+\sqrt{3}b \right)}^{3}}$.
$\therefore$ The result is $2\sqrt{2}{{a}^{3}}+3\sqrt{3}{{b}^{3}}+6\sqrt{3}{{a}^{2}}b+9\sqrt{2}a{{b}^{2}}={{\left( \sqrt{2}a+\sqrt{3}b \right)}^{3}}$.

Note: We should not confuse and make calculation mistakes while solving this problem. We can also find this result by assigning numerical values to the variable a and b. Whenever we get this type of problem, we should try to convert it into the combination of the sum of exponents of different terms to get the result. We can also solve this problem by using the binomial expansion of ${{\left( a+b \right)}^{n}}$. Similarly, we can expect problems with the powers of a and b greater than 3.