If a + b = 10 and ab = 16, find the value of


If $a+b=10$ and $a b=16$, find the value of $a^{2}-a b+b^{2}$ and $a^{2}+a b+b^{2}$



a + b = 10, ab = 16

We know that,

$(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$

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

$\Rightarrow a^{3}+b^{3}=(10)^{3}-3(16)(10)$

$\Rightarrow a^{3}+b^{3}=1000-480$

$\Rightarrow a^{3}+b^{3}=520$

Substitute, $a^{3}+b^{3}=520$

$a+b=10 i n a^{3}+b^{3}$

$=(a+b)\left(a^{2}+b^{2}-a b\right) a^{3}+b^{3}$

$=(a+b)\left(a^{2}+b^{2}-a b\right) 520$

$=10\left(a^{2}+b^{2}-a b\right) 520 / 10$

$=\left(a^{2}+b^{2}-a b\right)$

$\Rightarrow\left(a^{2}+b^{2}-a b\right)=52$

Now, we need to find $\left(a^{2}+b^{2}+a b\right)$

Add and subtract $2 a b$ in $a^{2}+b^{2}+a b$

$\Rightarrow a^{2}+b^{2}+a b-2 a b+2 a b$

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

Substitute a + b = 10, ab

$\Rightarrow a^{2}+b^{2}+a b=10^{2}-16=100-16=84$

Hence, the values of $\left(a^{2}+b^{2}-a b\right)=52$ and $\left(a^{2}+b^{2}+a b\right)=84$


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now