Advertisements
Advertisements
प्रश्न
Show that: (a − b)(a + b) + (b − c)(b + c) + (c − a)( c + a) = 0
Advertisements
उत्तर
\[\text { LHS } = \left( a - b \right)\left( a + b \right) + \left( b - c \right)\left( b + c \right) + \left( c + a \right)\left( c - a \right)\]
\[ = a^2 - b^2 + b^2 - c^2 + c^2 - a^2 \left[ \because \left( a + b \right)\left( a - b \right) = a^2 - b^2 \right]\]
\[ = a^2 - b^2 + b^2 - c^2 + c^2 - a^2 \]
\[ = 0\]
= RHS
Because LHS is equal to RHS, the given equation is verified.
संबंधित प्रश्न
Find each of the following product:
(−4x2) × (−6xy2) × (−3yz2)
Evaluate (2.3a5b2) × (1.2a2b2) when a = 1 and b = 0.5.
Find the following product: \[\frac{7}{5} x^2 y\left( \frac{3}{5}x y^2 + \frac{2}{5}x \right)\]
Find the following product: \[\frac{4}{3}a( a^2 + b^2 - 3 c^2 )\]
Multiply:
(2x + 8) by (x − 3)
Multiply:
(x2 + y2) by (3a + 2b)
(2xy + 3y2) (3y2 − 2)
Simplify:
(x3 − 2x2 + 5x − 7)(2x − 3)
Simplify:
(x2 − 3x + 2)(5x − 2) − (3x2 + 4x − 5)(2x − 1)
Simplify : (2.5p − 1.5q)2 − (1.5p − 2.5q)2
