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Using binomial theorem, evaluate ${{\left( 96 \right)}^{3}}$ .

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Last updated date: 22nd Mar 2024
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MVSAT 2024
Answer
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Hint: Start by writing 96 as 100 – 4 followed by the application of the binomial expansion of ${{\left( a-b \right)}^{3}}$ that you get by using the general expansion of ${{\left( a-b \right)}^{n}}$.

Complete step-by-step answer:
In the question we are asked to find the cube of 96 by the method of binomial expansion.
We know that 96 can be written as the difference of 100 and 4. On doing so, our expression becomes:
${{\left( 96 \right)}^{3}}={{\left( 100-4 \right)}^{3}}$
Now we know that the binomial expansion of ${{\left( a-b \right)}^{n}}$ , can be written as:
${{\left( a-b \right)}^{n}}={{\text{ }}^{n}}{{\text{C}}_{0}}{{a}^{n}}{{b}^{0}}{{-}^{n}}{{\text{C}}_{1}}{{a}^{n-1}}{{b}^{1}}{{+}^{n}}{{\text{C}}_{2}}{{a}^{n-2}}{{b}^{2}}-.........{{+}^{n}}{{\text{C}}_{n}}{{a}^{0}}{{\left( -b \right)}^{n}}$
Therefore, the binomial expansion of ${{\left( a-b \right)}^{3}}$ is:
${{\left( a-b \right)}^{3}}={{\text{ }}^{3}}{{\text{C}}_{0}}{{a}^{3}}{{b}^{0}}{{-}^{3}}{{\text{C}}_{1}}{{a}^{2}}{{b}^{1}}{{+}^{3}}{{\text{C}}_{2}}{{a}^{1}}{{b}^{2}}{{-}^{3}}{{\text{C}}_{3}}{{a}^{0}}{{b}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3{{b}^{2}}a$
Now using the formula for finding the value of ${{\left( 96 \right)}^{3}}={{\left( 100-4 \right)}^{3}}$ as asked in the question. On doing so, we get
${{\left( 100-4 \right)}^{3}}=\text{ 10}{{\text{0}}^{3}}-{{4}^{3}}-3\times {{100}^{2}}\times 4+3\times {{4}^{2}}\times 100$
$\Rightarrow {{\left( 100-4 \right)}^{3}}=\text{ 1000000}-64-120000+4800=884736$
Therefore, we can conclude that the value of ${{\left( 96 \right)}^{3}}$ comes out to be 884736.

Note: Always be careful with the signs that appear in the expansions, as the students are generally finding signs to be a concern while using the binomial expansions. Also, be careful about the calculation part, as in general cases, the questions involving binomial expansion have very long calculations, as we have seen in the above question. While breaking the number as a sum or difference of two numbers, try to ensure that one of the two numbers we are selecting is divisible by 10, as it reduces our effort of calculating the squares, cubes, or higher powers.