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| Cubic Impression (Posted on 2013-11-02) |
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Express each of X, Y and Z in terms of a,b,c,p,q and r so that the equation given below becomes an identity.
(a3+b3+c3 - 3abc)(p3+q3+r3 - 3pqr) = X3+Y3+Z3 - 3XYZ
Note: Disregard any permutations. For example, if (X, Y, Z) = (α, β, γ), then (X, Y, Z) = (β, γ, α) is invalid.
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Submitted by K Sengupta
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Rating: 5.0000 (2 votes)
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Solution:
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We know that:
a3+b3+c3 – 3abc = (a+b+c)(a2+b2+c2-ab-bc-ca)
Also, (a2+b2+c2-ab-bc-ca) can be further factorized as:
(a+bw+cw2) (a+bw2+cw) where w is the complex cube root of unity.
Similarly, p3+q3+r3 – 3abc = (p+q+r)(p+qw+rw2)(p+qw2+rw)
So, (a3+b3+c3 – 3abc)(p3+q3+r3 – 3pqr)
= (a+b+c) (a+bw+cw22)(a+b2+cw) (p+q+r)(p+qw+rw2)(p+qw2+rw)
Now, letting J = ap+br+cq, K = aq+bp+cr,, L = ar+bq+cp we see that:
(a+b+c) (p+q+r) = J+K+L
(a+bw+cw^2) (p+qw+rw2) = J+Kw+L2
(a+bw2+cw)(p+qw2+rw) = J+Kw2+Lw
So, (a3+b3+c3 – 3abc)(p3+q3+r3 – 3pqr)
= (J+K+L)(J+Kw+Lw2)( J+Kw2+Lw)
= J3+K3+L3 – 3JKL
Disregarding permutations, without any loss of generality, we can replace J, K and L respectively with X, Y and Z
So, disregarding permutations:
X = ap+br+cq
Y = aq+bp+cr
Z = ar+bq+cp
is the general solution for X, Y and Z in terms of a,b,c,p,q and r.
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