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Question: What is the formula for \({\left( {a + b} \right)^3}\)?...

What is the formula for (a+b)3{\left( {a + b} \right)^3}?

Explanation

Solution

The given question requires us to find the formula for (a+b)3{\left( {a + b} \right)^3}. It is a type of binomial expression. We can find the required binomial expansion by using the Binomial theorem. Binomial theorem helps us to expand the powers of binomial expressions easily and can be used to solve the given problem. We must know the formulae of combinations and factorials for solving the given question using the binomial theorem.

Complete step by step solution:
We have to find the formula for (a+b)3{\left( {a + b} \right)^3}. So, using the binomial theorem, the binomial expansion of (x+y)n{\left( {x + y} \right)^n} is ∑r=0n(nCr)(x)n−r(y)r\sum\nolimits_{r = 0}^n {\left( {^n{C_r}} \right){{\left( x \right)}^{n - r}}{{\left( y \right)}^r}} .
So, the binomial expansion of (a+b)3{\left( {a + b} \right)^3} is ∑r=03(3Cr)(a)3−r(b)r\sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( a \right)}^{3 - r}}{{\left( b \right)}^r}} .
Now, we have to expand the expression ∑r=03(3Cr)(a)3−r(b)r\sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( a \right)}^{3 - r}}{{\left( b \right)}^r}} and we are done with the formula for (a+b)3{\left( {a + b} \right)^3} .
∑r=03(3Cr)(a)3−r(b)r=(3C0)(a)3(b)0+(3C1)(a)2(b)1+(3C2)(a)1(b)2+(3C3)(a)0(b)3\sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( a \right)}^{3 - r}}{{\left( b \right)}^r}} = \left( {^3{C_0}} \right){\left( a \right)^3}{\left( b \right)^0} + \left( {^3{C_1}} \right){\left( a \right)^2}{\left( b \right)^1} + \left( {^3{C_2}} \right){\left( a \right)^1}{\left( b \right)^2} + \left( {^3{C_3}} \right){\left( a \right)^0}{\left( b \right)^3}
Since any number raised to power zero is one. So, evaluating the powers of a and b, we get,
Hence, ∑r=03(3Cr)(a)3−r(b)r=(3C0)a3+(3C1)a2b+(3C2)ab2+(3C3)b3\sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( a \right)}^{3 - r}}{{\left( b \right)}^r}} = \left( {^3{C_0}} \right){a^3} + \left( {^3{C_1}} \right){a^2}b + \left( {^3{C_2}} \right)a{b^2} + \left( {^3{C_3}} \right){b^3}
Substituting the values of combinations formulae, we get,
⇒∑r=03(3Cr)(a)3−r(b)r=(1)a3+(3)a2b+(3)ab2+(1)b3\Rightarrow \sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( a \right)}^{3 - r}}{{\left( b \right)}^r}} = \left( 1 \right){a^3} + \left( 3 \right){a^2}b + \left( 3 \right)a{b^2} + \left( 1 \right){b^3}
Simplifying the expression, we get,
⇒∑r=03(3Cr)(a)3−r(b)r=a3+3a2b+3ab2+b3\Rightarrow \sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( a \right)}^{3 - r}}{{\left( b \right)}^r}} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}
So, the formula for (a+b)3{\left( {a + b} \right)^3} is (a3+3a2b+3ab2+b3)\left( {{a^3} + 3{a^2}b + 3a{b^2} + {b^3}} \right).

Note:
The given problem can be solved by various methods. The easiest way is to apply the concepts of Binomial theorem as it is very effective in finding the binomial expansion. We need to remember the basic Combination formula to make the calculations easy.