# Evaluate the following using suitable identities:

(i) ${\left( {99} \right)^3}$ (ii) ${\left( {102} \right)^3}$ (iii) ${\left( {998} \right)^3}$

Last updated date: 27th Mar 2023

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Answer

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Hint: Try to break the number in terms of 10’s or 100’s.

For evaluating, we will be using these two identities:

$

{\left( {a - b} \right)^3} = {a^3} - {b^3} - 3ab\left( {a - b} \right){\text{ }} \ldots \left( 1 \right) \\

{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right){\text{ }} \ldots \left( 2 \right) \\

$

(i) ${\left( {99} \right)^3}$

$

\Rightarrow {\left( {100 - 1} \right)^3} \\

\Rightarrow {\left( {100} \right)^3} - {\left( 1 \right)^3} - 3\left( {100} \right)\left( 1 \right)\left( {100 - 1} \right){\text{ }}\left( {{\text{using }}\left( 1 \right)} \right) \\

\Rightarrow 1000000 - 1 - 300\left( {100 - 1} \right) \\

\Rightarrow 1000000 - 1 - 30000 + 300 \\

\Rightarrow 970299 \\

$

(ii) ${\left( {102} \right)^3}$

$

\Rightarrow {\left( {100 + 2} \right)^3} \\

\Rightarrow {\left( {100} \right)^3} + {\left( 2 \right)^3} + 3\left( {100} \right)\left( 2 \right)\left( {100 + 2} \right){\text{ }}\left( {{\text{using }}\left( 2 \right)} \right) \\

\Rightarrow 1000000 + 8 + 600\left( {100 + 2} \right) \\

\Rightarrow 1000000 + 8 + 60000 + 1200 \\

\Rightarrow 1061208 \\

$

(iii) ${\left( {998} \right)^3}$

$

\Rightarrow {\left( {1000 - 2} \right)^3} \\

\Rightarrow {\left( {1000} \right)^3} - {\left( 2 \right)^3} - 3\left( {1000} \right)\left( 2 \right)\left( {1000 - 2} \right){\text{ }}\left( {{\text{using }}\left( 1 \right)} \right) \\

\Rightarrow 1000000000 - 8 - 6000\left( {1000 - 2} \right) \\

\Rightarrow 1000000000 - 8 - 6000000 + 12000 \\

\Rightarrow 994011992 \\

$

Note: Whenever you see a large valued number has a power 2 or 3, always try to write that number in terms of 10’s or 100’s and then use square or cubic formulas. Because finding squares or cubes of 10 and 100 is an easier task.

For evaluating, we will be using these two identities:

$

{\left( {a - b} \right)^3} = {a^3} - {b^3} - 3ab\left( {a - b} \right){\text{ }} \ldots \left( 1 \right) \\

{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right){\text{ }} \ldots \left( 2 \right) \\

$

(i) ${\left( {99} \right)^3}$

$

\Rightarrow {\left( {100 - 1} \right)^3} \\

\Rightarrow {\left( {100} \right)^3} - {\left( 1 \right)^3} - 3\left( {100} \right)\left( 1 \right)\left( {100 - 1} \right){\text{ }}\left( {{\text{using }}\left( 1 \right)} \right) \\

\Rightarrow 1000000 - 1 - 300\left( {100 - 1} \right) \\

\Rightarrow 1000000 - 1 - 30000 + 300 \\

\Rightarrow 970299 \\

$

(ii) ${\left( {102} \right)^3}$

$

\Rightarrow {\left( {100 + 2} \right)^3} \\

\Rightarrow {\left( {100} \right)^3} + {\left( 2 \right)^3} + 3\left( {100} \right)\left( 2 \right)\left( {100 + 2} \right){\text{ }}\left( {{\text{using }}\left( 2 \right)} \right) \\

\Rightarrow 1000000 + 8 + 600\left( {100 + 2} \right) \\

\Rightarrow 1000000 + 8 + 60000 + 1200 \\

\Rightarrow 1061208 \\

$

(iii) ${\left( {998} \right)^3}$

$

\Rightarrow {\left( {1000 - 2} \right)^3} \\

\Rightarrow {\left( {1000} \right)^3} - {\left( 2 \right)^3} - 3\left( {1000} \right)\left( 2 \right)\left( {1000 - 2} \right){\text{ }}\left( {{\text{using }}\left( 1 \right)} \right) \\

\Rightarrow 1000000000 - 8 - 6000\left( {1000 - 2} \right) \\

\Rightarrow 1000000000 - 8 - 6000000 + 12000 \\

\Rightarrow 994011992 \\

$

Note: Whenever you see a large valued number has a power 2 or 3, always try to write that number in terms of 10’s or 100’s and then use square or cubic formulas. Because finding squares or cubes of 10 and 100 is an easier task.

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