greatest possible value of b is when a is zero.

so b^6 = 144
b^3 = 12 b = 2.xxx
b is between 3 and 4 so

Ans D

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The question states that \(b^6 = 144-a^6\). Hence for \(b\) to be maximum \(a=0\)
so \(b^6=144\) taking square root of both the sides, we get
\(b^3 = 12\). Now,
\(8<12<27\) or \(8<b^3<27\). Taking cube root of the inequality we get
\(2<b<3\). Hence \(b\) is definitely greater than \(2\) but less than \(3\) (to be precise; \(b^3 = 12\), or \(b= 2.289428\))

Ideally the greatest possible value of \(b\) is between \(2\) & \(3\), but we don't have any option stating that.
Hi Bunuel can you confirm whether Option D is correct and there are no typo errors

The greatest possible value of b turns out to be ~2.3, which is indeed between 2 and 3, but it will also be true to say that it's between -1,000 and 10,000 isn't it? Between x and y, here means in the range from x to y. So, we can say that 2.3 is between 2 and 4.
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Since we want to maximize the value of b, we want to minimize the value of a^6. Since a has an even power, the minimum value of a^6 is 0, which occurs when a = 0 (note: if a ≠ 0, then a^6 > 0). Thus, when a^6 = 0, we have:

b^6 = 144

Since 2^6 = 64 and 3^6 = 729, b is greater than 2 but less than 3.

Answer: D
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