Question

if a^1/3+b^1/3+c^1/3 so show that (a+b+c)^3= 3abc​

Answer 1

$\sf \large Question : - \\ \sf If \ {a}^{\frac{1}{3}}+ b^{\frac{1}{3}} + {c}^{ \frac{1}{3} } = 0 \ then \ (a+b+c)^3 = ?? \\ \\ \sf \Large ANSWER \\ \sf Given :- \\ \sf {a}^{\frac{1}{3}}+ b^{\frac{1}{3}} + {c}^{ \frac{1}{3} } = 0 \\ \\ \textsf{ Required to find :- }\\ \sf Value \ of \ (a+b+c)^3 = \ ?? \\ \\ \sf Solution : - \\ \sf Given \ that ; \\ \sf {a}^{\frac{1}{3}}+ b^{\frac{1}{3}} + {c}^{ \frac{1}{3} } = 0 \\ \sf Transposing \ c^{\frac{1}{3}} \ to \ right \ side \\ \sf {a}^{\frac{1}{3}}+ b^{\frac{1}{3}} = - c^{ \frac{1}{3} } (1)$

$\\ \sf Cubing \ on \ both \ sides \\ \sf \left( {a}^{\frac{1}{3}}+ b^{\frac{1}{3}} \right)^3= \left( - c^{ \frac{1}{3} } \right)^3 \\ \rm We \ know \ that; \\ \quad \bullet \tt \;(a^m)^n = a^{mn} \\ \quad \bullet \tt \; (a+b)^3 = a^3+b^3+3ab(a+b) \\ \\ \sf \left([ {a}^{\frac{1}{3}} ]^3 + [b^{\frac{1}{3}} ]^3 + 3 (a^{\frac{1}{3}} )( {b}^{ \frac{1}{3} })( {a}^{ \frac{1}{3} } + {b}^{ \frac{1}{3} } ) \right)= \big( - c^{ \frac{1}{3} } \big)^3 \\ \\ \sf \left( a + b + 3 (a^{\frac{1}{3}} )( {b}^{ \frac{1}{3} })( {a}^{ \frac{1}{3} } + {b}^{ \frac{1}{3} } ) \right)= \big( - c^{ \frac{1}{3} } \big)^3 \\ \sf su bstituting \ the \ value \ of \ a^{\frac{1}{3}}+b^{\frac{1}{3}} \ from \ eq-1 \\ \\ \tt \left( a + b + 3 (a^{\frac{1}{3}} )( {b}^{ \frac{1}{3} })( {c}^{ \frac{1}{3} } ) \right)= \big( - c^{ \frac{1}{3} } \big)^3 \\ \\ \sf \left( a+b+3 (a^{\frac{1}{3}} )( {b}^{ \frac{1}{3} })( {c}^{ \frac{1}{3} } ) \right)= \big( - c \big) \\ \\ \sf a+b+c = 3 (a^{\frac{1}{3}} )( {b}^{ \frac{1}{3} })( {c}^{ \frac{1}{3} } )$

$\\ \sf By \ cubing \ on \ both \ sides \\ \\ \sf (a+b+c)^3 = \left( 3 (a^{\frac{1}{3}} )( {b}^{ \frac{1}{3} })( {c}^{ \frac{1}{3} } ) \right)^3 \\ \\ \sf (a+b+c)^3 = \left( 27 (a^{\frac{3}{3}} )( {b}^{ \frac{3}{3} })( {c}^{ \frac{3}{3} } ) \right) \\ \\ \implies \underline{ \sf (a+b+c)^3 = 27abc } \\ \\ \\ \footnotesize{\sf{\red{\therefore \ (a+b+c)^3 = 27abc \ is \ the \ required \ answer }}}$