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# How do you expand ${{\left( 1-0.1 \right)}^{3}}$?

Last updated date: 14th Aug 2024
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Hint: Assume 1 as ‘a’ and 0.1 as ‘b’ and use the algebraic identity: - ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)$, to simplify the given expression. Substitute the assumed values of ‘a’ and ‘b’, calculate the answer by simple addition and subtraction.

Here, we have been provided with the expression ${{\left( 1-0.1 \right)}^{3}}$ and we have been asked to expand it. So, let us see how to calculate the value of this expression by expansion.
Now, as we can see that the exponent of the binomial term $\left( 1-0.1 \right)$ is 3, that means we have to multiply this term three times. So, assuming 1 as ‘a’ and 0.1 as ‘b’ we have,
$\Rightarrow {{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)$
Let us multiply the first two terms, so we have,
\begin{align} & \Rightarrow {{\left( a-b \right)}^{3}}=\left[ \left( a-b \right)\left( a-b \right) \right]\times \left( a-b \right) \\ & \Rightarrow {{\left( a-b \right)}^{3}}=\left[ {{a}^{2}}-ab-ba+{{b}^{2}} \right]\times \left( a-b \right) \\ \end{align}
Here, ab = ba, so we get,
$\Rightarrow {{\left( a-b \right)}^{3}}=\left[ {{a}^{2}}-2ab+{{b}^{2}} \right]\times \left( a-b \right)$
Now, let us multiply these two expressions in the R.H.S. to form a general expression, so we get,
\begin{align} & \Rightarrow {{\left( a-b \right)}^{3}}=\left( {{a}^{3}}-{{a}^{2}}b-2{{a}^{2}}b+2a{{b}^{2}}+{{b}^{2}}a-{{b}^{3}} \right) \\ & \Rightarrow {{\left( a-b \right)}^{3}}=\left( {{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}} \right) \\ \end{align}
So, here we have obtained the formula for the whole cube of the difference of two terms ‘a’ and ‘b’. Therefore, substituting back the initially assumed values of ‘a’ and ‘b’, we get,
$\Rightarrow {{\left( 1-0.1 \right)}^{3}}=\left( {{1}^{3}}-{{\left( 0.1 \right)}^{3}}-3\left( 0.1 \right){{\left( 1 \right)}^{2}}+3\left( 1 \right){{\left( 0.1 \right)}^{2}} \right)$
On simplification we get,
\begin{align} & \Rightarrow {{\left( 1-0.1 \right)}^{3}}=\left( 1-0.001-0.3+0.03 \right) \\ & \Rightarrow {{\left( 1-0.1 \right)}^{3}}=0.729 \\ \end{align}

Hence, the value of the given expression is 0.729.

Note: One may note that you can also get the answer by cubing 0.9 because 1 – 0.1 = 0.9. Note these algebraic identities are generally used to make our calculation easy. For example, if we have to calculate the value of ${{99}^{3}}$ then we can simplify it as $99=100-1$. This will make our calculations much easier because it is difficult to multiply 99 three times. You must remember other algebraic identities also like: - ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$, ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$ etc. because these are basic identities used in algebra.