Question

If $a+b=6$ and $a b=8,$ find the value of $a^{3}+b^{3}$

Answer 1

Hint: Use the formula of $a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)$ and $a^{2}+b^{2}=(a+b)^{2}-2 a b$ and put the given values.

Complete step-by-step answer:
Given, $a+b=6$ and $a b=8$
We know that $a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)$
Putting the given values, we get
$a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)$
$\Rightarrow a^{3}+b^{3}=(6)\left(a^{2}+b^{2}-8\right)$
Now, $a^{2}+b^{2}=(a+b)^{2}-2 a b=6^{2}-2 \times 8=36-16=20$
Putting the value in (i) we get
$a^{3}+b^{3}=(6)(20-8)=6 \times 12=72$
Hence, the required answer is 72 .

Note: To do this sum, one needs to be acquainted with the formulae of $a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)$ and $a^{2}+b^{2}=(a+b)^{2}-2 a b$ and understand the concept properly
Students go wrong in identifying the relation between the two given equations. Once the student realizes the relation between the given equations, the sum can be solved easily. A linear algebraic equation is nice and simple, containing only constants and variables to the first degree (no exponents or fancy stuff). To solve it, simply use multiplication, division, addition, and subtraction when necessary to isolate the variable and solve for " x ". This is the way to solve linear algebraic expressions.