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Algebra Level 3

If ∣ 1 + a x 1 + b x 1 + c x 1 + a 1 x 1 + b 1 x 1 + c 1 x 1 + a 2 x 1 + b 2 x 1 + c 2 x ∣ \left| \begin{array}{lcr}1+ax & 1+bx & 1+cx \\ 1+a_1x & 1+b_1x & 1+c_1x \\ 1+a_2x & 1+b_2x & 1+c_2x \end{array} \right| = A 0 + A 1 x + A 2 x 2 + A 3 x 3 \ A_0 + A_1x + A_2x^2 + A_3x^3 ,

Then A 0 A_0 is

NOTE : x ϵ \epsilon R \mathbb{R}

(Challenge-Solve it within 10 seconds)
abc 0 -1 1 4

Akhil Bansal by Akhil Bansal , 22, India

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2 solutions

Otto Bretscher
Sep 8, 2015

Setting x = 0 x=0 gives ∣ 1 1 1 1 1 1 1 1 1 ∣ = A 0 = 0 \left|\begin{array}{ccc}1&1&1\\1&1&1\\1&1&1\end{array}\right|=A_0=\boxed{0}

6 Helpful 0 Interesting 0 Brilliant 0 Confused
Pranjal Prashant
Sep 8, 2015

as all the constant will be cancelled out {in the form of 1{1-1}}, the answer is zero

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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