Question

If \[a + b = 8\] and \[ab = 6\] find the value of \[{a^3} + {b^3}\].

Answer 1

Hint:
Here the problem we are dealing with is a simple algebraic problem where it is given that \[a + b = 8\] and \[ab = 6\]. Now again, we know the formula of which is to be used in this problem, we simply substitute the given values and solve for \[{a^3} + {b^3}\].

Complete step by step solution:
Here, we are given, \[a + b = 8\] and \[ab = 6\]
Now, We also know that,
${\left({a+b}\right)}^{3}={a}^{3}+{b}^{3}+3ab{\left({a+b}\right)}$
Substituting the values \[a + b = 8\] and \[ab = 6\], we get,
$\Rightarrow {8}^{3}={a}^{3}+{b}^{3}+3 \times 6{\left(6\right)}$
On simplification we get,
$\Rightarrow {512}={a}^{3}+{b}^{3}+144$
On Changing sides, we get,
$\Rightarrow {512-144}={a}^{3}+{b}^{3}$
On further simplification we get,
$\Rightarrow {a}^{3}+{b}^{3}=368$

So, the value of \[{a^3} + {b^3}\] is 368 if \[a + b = 8\] and \[ab = 6\].

Note:
Derivation of expansion of \[{(a + b)^3}\] identity in algebraic approach.
\[ \Rightarrow {(a + b)^3} = ((a + b)(a + b)(a + b))\]
Solving for 2 terms first,
\[ \Rightarrow {(a + b)^3} = (a + b) \times ({a^2} + ab + ba + {b^2})\]
\[ \Rightarrow {(a + b)^3} = (a + b) \times ({a^2} + 2ab + {b^2})\]
Now, multiply the sum of two terms with the expansion of \[a + b\;\] whole square by the multiplication of algebraic expressions, we get,
\[ \Rightarrow {(a + b)^3} = a \times ({a^2} + 2ab + {b^2}) + b \times ({a^2} + 2ab + {b^2})\]
On further simplification we get,
\[ \Rightarrow {{\text{(a + b)}}^{\text{3}}}{\text{ = }}{{\text{a}}^3}{\text{ + 2}}{{\text{a}}^2}{\text{b + a}}{{\text{b}}^2}{\text{ + }}{{\text{a}}^2}{\text{b + 2a}}{{\text{b}}^2}{\text{ + }}{{\text{b}}^3}\]
\[ \Rightarrow {{\text{(a + b)}}^{\text{3}}}{\text{ = }}{{\text{a}}^3}{\text{ + 3}}{{\text{a}}^2}{\text{b + 3a}}{{\text{b}}^2}{\text{ + }}{{\text{b}}^3}\]
\[\therefore {{\text{(a + b)}}^{\text{3}}}{\text{ = }}{{\text{a}}^3}{\text{ + 3}}{{\text{a}}^2}{\text{b + 3a}}{{\text{b}}^2}{\text{ + }}{{\text{b}}^3}\]