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Source: — Data Sufficiency |
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by knight247 » Mon Sep 19, 2011 5:45 am
Let x³=a and 1/x³=b so a²=x^6 and b²=1/(x^6)

we need to find the value of a²+b²

1)a²-b²=128
(a-b)(a+b)=128 no way to solve further so INSUFFICIENT

2)a+b=14
Now way to solve further.

Combining both. Put a+b=14 in (a-b)(a+b)=128 we get (a-b)*14=128 so a-b=some value Solve this simultaneously with a+b=14. U'll get some value of a and b so you'll have the value of x^3 and 1/x^. Next find the value of x^6 and 1/x^6. Hence C
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by eshwarjayanth » Mon Sep 19, 2011 8:28 am
shankar.ashwin wrote:What is the value of (x^6+x^-6) ?

1) x^6 - x^-6 = 128
2) x^3 + x^-3 = 14
2 => x^3 + x^-3 = 14
squaring on both sides

=> (x^3)^2 + (x^-3)^2 + 2(x^3)(x^-3) = 196
=> x^6 + x^-6 = 194
So B is sufficient
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by shankar.ashwin » Mon Sep 19, 2011 10:45 am
The OA is D
shankar.ashwin wrote:What is the value of (x^6+x^-6) ?

1) x^6 - x^-6 = 128
2) x^3 + x^-3 = 14
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