nbsp;We have: nbsp; nbsp; nbsp; nbsp;1 b c2=12 b c2 #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Factors of Algebraic Expressions and Factorisation

Factorise:1-b...

Question

Factorise:
1 − (b − c)²

Open in App

Solution

We have:
$1 - {(b - c)}^{2} = {(1)}^{2} - {(b - c)}^{2} = \{1 + (b - c)\} \{1 - (b - c)\} = (1 + b - c) (1 - b + c)$

∴ $1 - {(b - c)}^{2} = (1 + b - c) (1 - b + c)$

Suggest Corrections

Similar questions

Simplify and factorise (a + b + c)² - (a - b - c)² + 4b² - 4c²

a^{2} - (b - c)^{2}

125 a^{3} + b^{3} + 64 c^{3} - 60 a b c

a^{3} + b^{3} + c^{3} - 3 a b c = \frac{1}{2} (a + b + c) [{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}]

1^{3} + b^{3} + 8 c^{3} - 6 b c

a^{3} + b^{3} + c^{3} - 3 a b c = \frac{1}{2} (a + b + c) [{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}]

125 - 8 x^{3} - 27 y^{3} - 90 x y

a^{3} + b^{3} + c^{3} - 3 a b c = \frac{1}{2} (a + b + c) [{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}]

Join BYJU'S Learning Program