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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