Add $a^{2} + 8 a b$ and $b^{2} - 6 a b$ .

Open in App

Solution

If we add Add $a^{2} + 8 a b$ and $b^{2} - 6 a b$ we get,
$(a^{2} + 8 a b) + (b^{2} - 6 a b)$
$= a^{2} + 2 a b + b^{2}$
$= (a + b)^{2}$ .

Suggest Corrections

Similar questions

4 a^{2} b - 2 b^{2} + 3 c^{2} a + 4 a b c 5

a^{2} b + b^{2} - 2 c^{2} + 8 a b

[\frac{a^{2} - 5 a b}{a^{2} - 6 a b + 5 b^{2}} \times \frac{a^{2} - b^{2}}{a^{2} + a b}]

sin θ + sin ϕ = a

cos θ + cos ϕ = b

(a \neq b, a \neq 0, b \neq 0)

a = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}

b = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}

a^{2} + b^{2} - 6 a b

Join BYJU'S Learning Program