Question

$Withoutactuallycalculatingthecubesfindthevalueof{\left(8\right)}^3+({7)}^3+({5)}^{3}and{\left(2\right)}^3+{\left(3\right)}^3+{(-3)}^3.$

Answer 1

given, $8^3+7^3+5^3$

${(a+b+c)}^3=a^3+b^3+c^3+3[(a+b+c)(ab+bc+ca)-abc]$

$\Rightarrow {(8+7+5)}^3=8^3+7^3+5^3+3[(8+7+5)(8\cdot 7+7\cdot 5+5\cdot 8)-8\times 7\times 5]$

$\Rightarrow {\left(20\right)}^3=8^3+7^3+5^3+3[\left(20\right)\times 131-280]$

$\Rightarrow 8^3+7^3+5^3=8000-7020=980$

Now,

$2^3+3^3+(-3{)}^3$

$=2^3+3^3-3^3$

$=2^3=8$