Question : Let $\sqrt[3]{a}=\sqrt[3]{26}+\sqrt[3]{7}+\sqrt[3]{63}$. Then:
Option 1: $a<729$ but $a>216$
Option 2: $a<216$
Option 3: $a>729$
Option 4: $a = 729$
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Correct Answer: $a<729$ but $a>216$
Solution :
$\sqrt[3]{a}=\sqrt[3]{26}+\sqrt[3]{7}+\sqrt[3]{63}$
As we know, $\sqrt[3]{26}< 3$, $\sqrt[3]{7}< 2$, and $\sqrt[3]{63}< 4$
So, $\sqrt[3]{a}=\sqrt[3]{26}+\sqrt[3]{7}+\sqrt[3]{63}$
⇒ $\sqrt[3]{a} < 3+2+4$
⇒ $\sqrt[3]{a}<9$
⇒ $a<729$
And,
$\sqrt[3]{a}=\sqrt[3]{26}+\sqrt[3]{7}+\sqrt[3]{63}$
⇒$\sqrt[3]{a}>2 + 1 + 3$
⇒ $a>216$
So, $a>216$
Hence, the correct answer is $a<729$ but $a>216$.
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