Question

What is $(a^3-b^3)$ if $ab=6$ and $(a+b)=5$ and $a>b$?

• A. 18
• B. 19
• C. 27
• D. 8

Answer 1

The correct option is B

19

$Givenab=6and(a+b)=5$

$\Rightarrow (a-b{)}^3=a^3-b^3-3ab(a-b)$

We need to find (a - b)

$\Rightarrow (a-b{)}^2=a^2+b^2-2ab$

$\Rightarrow (a-b{)}^2=a^2+b^2-2(6)$

$\Rightarrow (a-b{)}^2=a^2+b^2-12$

Now we need to find $(a^2+b^2)$

$\Rightarrow (a+b{)}^2=a^2+b^2+2ab$

$\Rightarrow (5{)}^2=a^2+b^2+2(6)$

$\Rightarrow 25=a^2+b^2+12$

$\Rightarrow (a^2+b^2)=25-12=13$

Substituting this value in $(a-b{)}^2=a^2+b^2-12$, we get

$(a-b{)}^2=13-12=1$

$(a-b)=\pm 1$

$\Rightarrow (a-b)=1\because Given(a>b)$

Now substituting (a - b) = 1

$\Rightarrow (a-b{)}^3=a^3-b^3-3ab(a-b)$

$\Rightarrow (1{)}^3=(a^3-b^3)-3(6\left)\right(1)$

$\Rightarrow 1=(a^3-b^3)-18$

$\Rightarrow (a^3-b^3)=18+1=19$