Answer

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**Hint:**In order to solve this problem we need to convert the given expression in $({a^2} - {b^2})$. Then we have to use the identity ${(a + b)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}$. Doing this and solving algebraically will give you the right answer.

**Complete step-by-step answer:**

In this question we have to solve the given expression, we have given

$

{\left( {\sqrt 2 + 1} \right)^6} - {(\sqrt 2 - 1)^6} \\

\Rightarrow {\left( {{{\left( {\sqrt 2 + 1} \right)}^3}} \right)^2} - {\left( {{{\left( {\sqrt 2 - 1} \right)}^3}} \right)^2} \\

$

And it is in the form of $({a^2} - {b^2}) = (a + b)(a - b)$

And hence on applying the same formulae in above equation, we have

$ \Rightarrow \left[ {{{\left( {\sqrt 2 + 1} \right)}^3} + {{\left( {\sqrt 2 - 1} \right)}^3}} \right]\left[ {\left[ {{{\left( {\sqrt 2 + 1} \right)}^3} - {{\left( {\sqrt 2 - 1} \right)}^3}} \right]} \right]$

Now we know that,

$\left[ {\because {{(a + b)}^3} = (a + b)({a^2} - ab + {b^2})} \right]$ and $[\because {(a - b)^3} = (a - b)({a^2} + ab + {b^2})]$

And hence on applying the above formula we have,

$ \Rightarrow \left[ {{{\left( {\sqrt 2 + 1} \right)}^3} + {{\left( {\sqrt 2 - 1} \right)}^3}} \right]\left[ {\left[ {{{\left( {\sqrt 2 + 1} \right)}^3} - {{\left( {\sqrt 2 - 1} \right)}^3}} \right]} \right]$=$\left[ {2\sqrt 2 + 6 + 3\sqrt 2 + 1 + 2\sqrt 2 - 6 + 3\sqrt 2 - 1} \right]\left[ {2\sqrt 2 + 6 + 3\sqrt 2 + 12\sqrt 2 - 6 - 3\sqrt 2 - 1} \right]$

And hence on solving, we have

$

\Rightarrow 10\sqrt 2 \times 14 \\

\Rightarrow 140\sqrt 2 \\

$

And hence $140\sqrt 2 $will be the solution to our question.

**Note:**In this type of question first of all we have to convert the given expression in the form of $({a^2} - {b^2})$ and then on simplification we have to apply ${a^3} + {b^3}$and ${a^3} - {b^3}$ hence on solving we can have our answer. In such problems we just need to remember the formulas and their conversion and need to know how the equation can be converted into an identity. Knowing this will solve your problems and will give you the right answer.

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