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Question

Show that ∣ ∣1+a1111+b1111+c∣ ∣=abc(1+1a+1b+1c)=abc+bc+ca+ab.

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Solution

∣ ∣1+a1111+b1111+c∣ ∣

dividing R1,R2,R3 by a,b,c respectively.

=abc∣ ∣ ∣ ∣ ∣ ∣1+1a1a1a1b1+1b1b1c1c1+1c∣ ∣ ∣ ∣ ∣ ∣

R1R1+R2+R3

=abc∣ ∣ ∣ ∣ ∣ ∣1+1a+1b+1c1+1a+1b+1c1+1a+1b+1c1b1+1b1b1c1c1+1c∣ ∣ ∣ ∣ ∣ ∣

Taking out common 1+1a+1b+1c from R1

=abc(1+1a+1b+1c)∣ ∣ ∣ ∣ ∣1111b1+1b1b1c1c1+1c∣ ∣ ∣ ∣ ∣

C2C2C1

C3C3C1

=abc(1+1a+1b+1c)∣ ∣ ∣ ∣ ∣1001b101c01∣ ∣ ∣ ∣ ∣

expanding along R1

=abc(1+1a+1b+1c)=abc+bc+ca+ab




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