First slide

Binomial theorem for positive integral Index

Question

If $a =^{20} C_{0} +^{20} C_{3} +^{20} C_{6} +^{20} C_{9} + \dots; b =^{20} C_{1} +^{20} C_{4} +^{20} C_{7} + \dots;$ and $c =^{20} C_{2} +^{20} C_{5} +^{20} C_{8} + \dots,$ then

Moderate

Question

Value of $a^{3} + b^{3} + c^{3} - 3 abc$ is

A

2¹⁰
B

2²⁰
C

0
D

2¹⁹

Solution

We have
$\begin{array}{l} a + b + c = 2^{20} \\ Now, a^{3} + b^{3} + c^{3} - 3 abc \\ = (a + b + c) (a + bω + {cω}^{2}) (a + {bω}^{2} + cω) \\ = 2^{20} (1 + ω)^{20} {(1 + ω^{2})}^{20} \\ = 2^{20} \end{array}$

Question

Value of $(a - b)^{2} + (b - c)^{2} + (c - a)^{2}$ is

A

1
B

2
C

2²⁰
D

2⁴⁰

Solution

$\begin{matrix} (a - b)^{2} + (b - c)^{2} + (c - a)^{2} \\ = 2 (a + bω + {cω}^{2}) (a + {bω}^{2} + cω) \\ = 2 \end{matrix}$

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