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Question

∣ ∣x+42x2x2xx+42x2x2xx+4∣ ∣=(5x+4)(4x)2.

∣ ∣y+kyyyy+kyyyy+k∣ ∣=k2(3y+k)

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Solution

LHS=∣ ∣x+42x2x2xx+42x2x2xx+4∣ ∣=∣ ∣5x+42x2x5x+4x+42x5x+42xx+4∣ ∣ (using C1C1+C2+C3)

=(5x+4)∣ ∣12x2x1x+42x12xx+4∣ ∣ [take out (5x+4)common from C1].

=(5x+4)∣ ∣12x2x0x+4000x+4∣ ∣
Expanding along C1, we get
=(5x+4)[1(4x)(4x)]=(5x+4)(4x2)=RHS

LHS=∣ ∣y+kyyyy+kyyyy+k∣ ∣=k2(3y+k)=∣ ∣3y+kyy3y+ky+ky3y+kyy+k∣ ∣ (using C1C1+C2+C3)

=(3y+k)∣ ∣1yy1y+ky1yy+k∣ ∣ [take out (3y+k) common from C1]

=3y+k∣ ∣1yy0k000k∣ ∣ (using R2R2R1 and R3R3R1)
Expanding along C3. we get
=(3y+k)(1×k.k)=k2(3y+k)=RHS.


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