Question

without actually cubing , find cubes of 48³ - 30³ - 18³

Answer 1

Answer :


${48}^{3} - {30}^{3} - {18}^{3} \\ \\here \: we \: need \: to \: find \: cubes \: without \: actual \: calculations \\ \\ now \: as \: we \: know \: that \\ \\ {a}^{3} + {b}^{3} + {c}^{3} = 3abc \\ if \: a + b + c = 0 \\ \\ let \: us \: assume \: that \\ here \: a = {48}^{3} \\ b = - 30 {}^{3} \\ c = { - 18}^{3} \\ \\ therefore \\ \\ 4 {8}^{3} + ( - 30 {)}^{3} + ( - 18 {)}^{3} = 3(48) ( - 30)( - 18) \\ if \: 48 + ( - 30) + ( - 18) = 0 \\ \\ 48 - 30 - 18 = 0 \\ \\ 0 = 0 \\ \\ now \: 48 + ( - 30) + ( - 18) = 0 \\ \\ therefore \\ \\ (4 {8)}^{3} + ( - 3 {0)}^{3} + ( - 18 {)}^{3} = 3(48)( - 30)( - 18)$


$3 \times 48 ( - 30)( - 18) \\ \\ 3 \times 48 \times30 \times 18 \\ \\ 90 \times 48 \times 18 \\ \\ 4320 \times 18 \\ \\ 77760$

$therefore \\ \\ 48 {}^{3} - 30 {}^{3} - 1 {8}^{3} = 77760$



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