Question

please solve this fast, will mark as brainliest !!!

Answer 1

✍✍HERE IS YOUR ANSWER..⬇⬇
================================

$\underline{ANSWER}$ ➡

$\boxed{(1) \: 2 \sqrt[3]{a} - 4 \sqrt[3]{b} - \sqrt[3]{c} = 0}$

$If \: \: 8a - 64b - c = 24 \sqrt[3]{abc} \: \:, where \: \: \\ a \: ,\: b \: ,\: c \: \: ≠ \: 0 \:. \: Then \\ \\ (1) \: \: 2 \sqrt[ 3]{a} - 4 \sqrt[3]{b} - \sqrt[3]{c} = 0 \: \: \: can \: \: be \: \: true \: .$

$\underline{SOLUTION}$ ➡

$We \: \: know \: \: that, \\ \\ If \: \: x + y + z = 0 \: \: Then ,\\ \\ x {}^{3} + y {}^{3} + z {}^{3} = 3 xyz$

$Here ,\\ \\ 2 \sqrt[3]{a} - 4 \sqrt[3]{b} - \sqrt[3]{c} = 0 \\ \\ = > 2(a) {}^{ \frac{1}{3} } + [ - 4(b) {}^{ \frac{1}{3} } ] + ( -c ) {}^{ \frac{1}{3} } = 0$

$Let, \\ \\ 2(a) {}^{ \frac{1}{3} } = x \: \: , \: \: [ - 4(b) {}^{ \frac{1}{3} } ]= y \: \: and \: \: ( -c ) {}^{ \frac{1}{3} } = z$

$So ,\\ \\ x + y + z = 0 \\ \\ Therefore, \\ \\ x {}^{3} + y {}^{3} + z {}^{3} = 3xyz$

$So ,\\ \\ \:[2(a) {}^{ \frac{1}{3} }] {}^{3} + [- 4(b) {}^{ \frac{1}{3} } ] {}^{3} + [( -c ) {}^{ \frac{1}{3} } ] {}^{3} \\ \\ = 3 \times 2(a) {}^{ \frac{1}{3} } \times [- 4(b) {}^{ \frac{1}{3} } ] \times ( -c ) {}^{ \frac{1}{3} } \\ \\ => (2){}^{3}a + (-4){}^{3}b+(-c) = 24 (abc){}^{\frac{1}{3} } \\ \\ = > 8a - 64b - c = 24 \sqrt[3]{abc}$

_______________________________

✝✝…HOPE…IT…HELPS…YOU…✝✝