Question
If abc = 1, then find the value of:
11+a+b−1+11+b+c−1+11+c+a−1 [4 MARKS]
11+a+b−1+11+b+c−1+11+c+a−1 [4 MARKS]
Answer:
:
Formula: 1 Mark
Steps: 2 Marks
Answer: 1 Mark
Given that,
abc=1
So, c=1÷ab=(ab)−1
Similarly,
a=(bc)−1
b=(ac)−1
And,
ab=c−1
ac=b−1
bc=a−1
11+a+b−1+11+b+c−1+11+c+a−1
=11+a+b−1+b−1b−1+bb−1+b−1c−1+aa+ac+aa−1
[Multiplying 2nd term by b−1b−1 and 3rd term by aa]
=11+a+b−1+b−1b−1+1+a+aa+b−1+1
∵aman=am−n
=1+b−1+a1+a+b−1
=1
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:
Formula: 1 Mark
Steps: 2 Marks
Answer: 1 Mark
Given that,
abc=1
So, c=1÷ab=(ab)−1
Similarly,
a=(bc)−1
b=(ac)−1
And,
ab=c−1
ac=b−1
bc=a−1
11+a+b−1+11+b+c−1+11+c+a−1
=11+a+b−1+b−1b−1+bb−1+b−1c−1+aa+ac+aa−1
[Multiplying 2nd term by b−1b−1 and 3rd term by aa]
=11+a+b−1+b−1b−1+1+a+aa+b−1+1
∵aman=am−n
=1+b−1+a1+a+b−1
=1
Was this answer helpful ?
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