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Question

If ∣ ∣xx+yx+y+z2x3x+2y4x+3y+2z3x6x=3y10x+6y+3z∣ ∣=64, then the real value of x is .........

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Solution

=x∣ ∣1x+yx+y+z23x+2y4x+3y+2z36x+3y10x+6y+3z∣ ∣
Using C3C3zC1C2C2yC1
=x2∣ ∣11x+y234x+3y3210x+6y∣ ∣
=x3∣ ∣1112343610∣ ∣[C3C3yC2]=x3(68+3)=x3
Given x3=64x=4,4ω,4ω2

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