Q6 of 274 Page 91

Expand: (3a + 4b + 5c)2.

We have,

(3a + 4b + 5c)2


We know that,


(a + b + c)2 = a2 + y2 + c2 + 2 (xy + yz + zx)


∴ (3a + 4b + 5c)2 = (9a2 + 16b2 + 25c2 + 2 × 3a × 4b + 2 × 4b × 5c + 2 × 5c × 3a)


= 9a2 + 16b2 + 25c2 + 24ab + 40bc + 30ac


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