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x3/ x3+27+9x(x+3) = x-3 / x+6
We are given that x3/(x+3)3 = x-3/(x+6). From componendo-dividendo, we get (x+3)3-x3/(x+3)3+x3 = (x+6) - (x-3)/(x+6) + (x-3) or 3 [x2+(x+3)2 + x(x+3)]/(2x+3)(x2+(x+3)2-x(x+3)] = 9/(2x+3) So, either x = -3/2 or 3[x2+(x+3)2-x(x+3)] = x2+(x+3)2 + x(x+3) The second equation would imply 9 = 0 and hence x = -3/2 is the only possibility
We are given that x3/(x+3)3 = x-3/(x+6). From componendo-dividendo, we get
(x+3)3-x3/(x+3)3+x3 = (x+6) - (x-3)/(x+6) + (x-3)
or 3 [x2+(x+3)2 + x(x+3)]/(2x+3)(x2+(x+3)2-x(x+3)] = 9/(2x+3)
So, either x = -3/2 or 3[x2+(x+3)2-x(x+3)] = x2+(x+3)2 + x(x+3)
The second equation would imply 9 = 0 and hence x = -3/2 is the only possibility
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