Advertisements
Advertisements
प्रश्न
Verify the following:
(a + b + c)(a2 + b2 + c2 – ab – bc – ca) = a3 + b3 + c3 – 3abc
Advertisements
उत्तर
Taking L.H.S. = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
= a(a2 + b2 + c2 – ab – bc – ca) + b(a2 + b2 + c2 – ab – bc – ca) + c(a2 + b2 + c2 – ab – bc – ca) ...[Distributive law]
= a3 + ab2 + ac2 – a2b – abc – a2c + ba2 + b3 + bc2 – b2a – b2c – bca + ca2 + cb2 + c3 – cab – c2b – c2a
= a3 + b3 + c3 – 3abc
= R.H.S.
Hence verified.
APPEARS IN
संबंधित प्रश्न
Expand: (3x + 4y)(3x - 4y)
(7x + 3)(7x – 4) = 49x2 – 7x – 12
Factorise: 4x2 – 9y2
Simplify using identities
(3p + q)(3p – q)
Using suitable identities, evaluate the following.
9.8 × 10.2
Using suitable identities, evaluate the following.
(132)2 – (68)2
Factorise the following using the identity a2 – b2 = (a + b)(a – b).
x4 – y4
Factorise the following using the identity a2 – b2 = (a + b)(a – b).
x4 – y4 + x2 – y2
Verify the following:
(ab + bc)(ab – bc) + (bc + ca)(bc – ca) + (ca + ab)(ca – ab) = 0
The product of two expressions is x5 + x3 + x. If one of them is x2 + x + 1, find the other.
