If a be one A.M and G1 and G2 be the G.Ms between b and c then G31+G32=
A
abc
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B
2abc
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C
3abc
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D
4abc
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Solution
The correct option is B2abc Lets consider b=p3,c=q3 G1=p2q G2=pq2 (G1)3+(G2)3=(pq)3[p3+q3] =bc(b+c) a=(b+c)/2 2a=b+c So the expression for (G1)3+(G2)3=2abc Option B