Question

Without actually calculating the cubes, find the value of \[{{45}^{3}}-{{25}^{3}}-{{20}^{3}}\]

Answer 1

Hint: We can solve this question by using formula of difference of two cubes as\[{{a}^{3}}-{{b}^{3}}-(a-b)({{a}^{2}}+{{b}^{2}}+ab)\] and also difference of square formula ${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$
Complete step by step solution:
Given expression is
\[{{45}^{3}}-{{25}^{3}}-{{20}^{3}}\]
In first two term we can apply difference of two cubes formula \[[{{a}^{3}}-{{b}^{3}}-(a-b)({{a}^{2}}+{{b}^{2}}+ab)]\]
\[\Rightarrow (45-25)[{{45}^{2}}+{{25}^{2}}+(45\times 25)]-{{20}^{3}}\]
\[\Rightarrow 20[{{45}^{2}}+{{25}^{2}}+(45\times 25)]-{{20}^{3}}\]
Now we can take 20 as common
\[\Rightarrow 20[{{45}^{2}}+{{25}^{2}}+(45\times 25)-{{20}^{2}}]\]
Now take \[{{25}^{2}}-{{20}^{2}}\]as \[{{a}^{2}}-{{b}^{2}}\]
\[\therefore [{{a}^{2}}-{{b}^{2}}=(a+b)(a-b)]\]
\[\Rightarrow 20[{{45}^{2}}+(45\times 25)+(25+20)(25-20)]\]
\[\Rightarrow 20[{{45}^{2}}+(45\times 25)+(45\times 5)]\]
Now we can take 45 as common
\[\Rightarrow 20[45(45+25+5)]\]
\[\Rightarrow 20[45(75)]\]
\[\Rightarrow 20\times 45\times 75\]
$\Rightarrow 67500$
Note: Some children do mistake in formulae like instead of ‘+’ sign they use ‘-’ which later on spoils the complete solution. Therefore, this should be kept in mind.
Moreover, these simplification questions seem easy but we should keep in mind every step and side by side recheck every step for the accuracy of the solution. Trick was selecting 25 and 20 for (a+b)(a-b) identity instead of 45 and 20.