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Question

Shew that b3c3+c3a3+a3b3=5a2b2c2 is the eliminant of ax+yz=bc,by+zx=ca,cz+xy=ab,xyz=abc.

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Solution

Substituting z=abcxy in first three equations,
we have,
ax2bcx+abc=0
by2cay+abc=0
x2y2abxy+abc2=0
From the last two equations we have
y2a2bc(c2bx)=yabc(x2bc)1ax(b2cx)
Hence, a3bcx(c2bx)(b2cx)=a2b2c2(x2bc)2
i.e., bcx4abcx3+x2(ab3+ac32b2c2)ab2c2x+b2c2=0
It remains to eliminate x between this equation and ax3bcx+abc=0
Multiply the first equation by 'a' and the equation ax2bcx+abc=0 by bc(bca2)x and subtract, then
(a3b3+a3c3+b3c34a2b2c2)x2ab3c3x+a2b3c3=0
Multiply ax2bcx+abc=0 by
ab2c2 and subtract from this last equation;-
(a3b3+b3c3+c3a35a2b2c2)x2=0
a3b3+b3c3+c3a3=5a2b2c2

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