Question 16

$a^{2} - b^{2}$ is equal to
a) $(a - b)^{2}$
b) $(a - b) (a - b)$
c) $(a + b) (a - b)$
d) $(a + b) (a + b)$

Open in App

Solution

c) $(a + b) (a - b)$

a) $(a - b)^{2} = a^{2} - 2 a b + b^{2}$
b) $(a - b) (a - b) = (a - b)^{2} = a^{2} - 2 a b + b^{2}$
c) $(a + b) (a - b) = a (a - b) + b (a - b)$
$= a^{2} - a b + b a - b^{2}$ $= a^{2} - b^{2}$ $[∵ a b = b a]$
d) $(a + b) (a + b) = (a + b)^{2} = a^{2} + 2 a b + b^{2}$

Suggest Corrections

Similar questions

a^{5} b - a b^{5}

\tan θ = \frac{a}{b}, then \frac{a \sin θ + b \cos θ}{a \sin θ - b \cos θ}

\frac{a^{2} + b^{2}}{a^{2} - b^{2}}

\frac{a^{2} - b^{2}}{a^{2} + b^{2}}

\frac{a + b}{a - b}

\frac{a - b}{a + b}

Explore more