Question

Evaluate each of the following using suitable identities:
A. ${{\left( 104 \right)}^{3}}$
B. ${{\left( 999 \right)}^{3}}$

Answer 1

Hint: We first find the simplification of the given polynomial ${{\left( 104 \right)}^{3}}$ and ${{\left( 999 \right)}^{3}}$ according to the identity ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$ and ${{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}$. We need to simplify the cubic polynomial of sum or difference of two numbers. We solve the multiplication to find the simplified form of ${{\left( 104 \right)}^{3}}$ by replacing with $a=100;b=4$ and ${{\left( 999 \right)}^{3}}$ by replacing with $a=1000;b=1$.

Complete step-by-step answer:
We need to find the simplified form of ${{\left( 104 \right)}^{3}}$ and ${{\left( 999 \right)}^{3}}$.
We are going to use the identity ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$ and ${{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}$.
I.We express ${{\left( 104 \right)}^{3}}$ as the cube of the sum of two numbers. We take $a=100;b=4$ for the identity of ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$
${{\left( 104 \right)}^{3}}={{\left( 100+4 \right)}^{3}}={{100}^{3}}+3\times {{100}^{2}}\times 4+3\times 100\times {{4}^{2}}+{{4}^{3}}$
Therefore, the simplified form of ${{\left( 104 \right)}^{3}}$ is
${{\left( 100+4 \right)}^{3}}=1000000+120000+4800+64=1124864$

II.We express ${{\left( 999 \right)}^{3}}$ as the cube of difference of two numbers. We take $a=1000;b=1$ for the identity of ${{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}$
${{\left( 999 \right)}^{3}}={{\left( 1000-1 \right)}^{3}}={{1000}^{3}}-3\times {{1000}^{2}}\times 1+3\times 1000\times {{1}^{2}}-{{1}^{3}}$
Therefore, the simplified form of ${{\left( 999 \right)}^{3}}$ is
${{\left( 1000-1 \right)}^{3}}=1000000000-3000000+3000-1=997002999$.


Note: We also can use the binomial theorem to find the general form and then put the value of 3. We have ${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}{{b}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+....+{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+....+{}^{n}{{C}_{n}}{{a}^{0}}{{b}^{n}}$. We need to find the cube of the sum of two numbers. So, we put $n=3$.
${{\left( a+b \right)}^{3}}={}^{3}{{C}_{0}}{{a}^{3}}{{b}^{0}}+{}^{3}{{C}_{1}}{{a}^{3-1}}{{b}^{1}}+{}^{3}{{C}_{2}}{{a}^{3-2}}{{b}^{2}}+{}^{3}{{C}_{3}}{{a}^{3-3}}{{b}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$.
In this way we also simplify the term of ${{\left( a-b \right)}^{3}}$.