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Question

If a, b and c are distinct, solve the equation ∣ ∣ ∣x2a2x2b2x2c2(xa)3(xb)3(xc)3(x+a)3(x+b)3(x+c)3∣ ∣ ∣=0
How many distinct solutions does this have?

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Solution

1x∣ ∣ ∣x3a2xx3b2xx3c2xx3a33x2+3a2xx3b33x2+3b2xx3c33x2c+3c2xx3+a33x2a+3a2xx3+b3+3x2b+3xb2x3+c3+3x2c+3cx2∣ ∣ ∣R2R2+R31x∣ ∣ ∣x3a2x(x3b2x)x3c2x2x3+6a22x3+6b2x2x3+6c2x(x+a)3(x+b)3(x+c)3∣ ∣ ∣=0R12R1R31x∣ ∣ ∣8a2x8b2x8c2x2x3+6a2x2x3+6b2x2x3+6c2x(x+a)3(x+b)3(x+c)3∣ ∣ ∣=0R2R2+68R11x∣ ∣ ∣8a2x8b2x8c2x2x32x32x3(x+a)3(x+6)3(x+c)3∣ ∣ ∣=0R3R3(R2+R1)16x∣ ∣ ∣a2xb2xc2xx3x3x3a3+3x2ab3+3x2bc3+3x2c∣ ∣ ∣=016x4x∣ ∣ ∣a2b2c2111a2+3x2ab3+3x2bc3+3x2c∣ ∣ ∣=0c1c1c2c2+c2c316x3∣ ∣ ∣a2b2(b2c)2c2001(3x2+a2+b2+ab(3x2+b2+c2cb)c3+3x2c∣ ∣ ∣=0,(ab),(bc)16(ab)(bc)x3∣ ∣ ∣(a+b)(b2+c2c2001(3x2+a2+b2+ab)3x2+b2+c2+cb)(+3x2c)∣ ∣ ∣=0R3R3R2116(ab)(bc)x3∣ ∣ ∣(a+b)(b+c)c20013x2ab(3x2cb)(c2+3x2c)c4∣ ∣ ∣=0+16(ab)(bc)x3[(a+b)(3x2cb)(3x2ab)(b+c)]=016(ab)(bc)x3[3x2[(a+bbc)]]+[abccb2+ab2+abc]16(a6)(bc)x3(3x2(ac)+(ac)b2)16(ab)(bc)(ac)x3(3x2+62)=0

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