If [tex]a^{1 / 3}+b^{1 / 3}+c^{1 / 3}=0[/tex] then:
a) [tex]a+b+c=0[/tex]
b) [tex](a+b+c)^3=27 a b c[/tex]
c) [tex]a+b+c=3(a b c)^{1 / 3}[/tex]
d) [tex]a^3+b^3+c^3=0[/tex]
Answer (2)
The two correct results from the equation a 1/3 + b 1/3 + c 1/3 = 0 are a + b + c = 3 ( ab c ) 1/3 and ( a + b + c ) 3 = 27 ab c . The options stating that a + b + c = 0 and a 3 + b 3 + c 3 = 0 are not necessarily valid. Therefore, the correct answers are (b) and (c).
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Given a 1/3 + b 1/3 + c 1/3 = 0 , substitute x = a 1/3 , y = b 1/3 , and z = c 1/3 , so x + y + z = 0 .
From x + y = − z , derive ( x + y ) 3 = ( − z ) 3 , leading to x 3 + y 3 + z 3 = 3 x yz .
Substitute back to obtain a + b + c = 3 ( ab c ) 1/3 .
Cube both sides to find ( a + b + c ) 3 = 27 ab c . Thus, the correct options are ( a + b + c ) 3 = 27 ab c and a + b + c = 3 ( ab c ) 1/3 .
Explanation
Understanding the Problem We are given that a 1/3 + b 1/3 + c 1/3 = 0 . We need to determine which of the given options is correct.
Substitution Let x = a 1/3 , y = b 1/3 , and z = c 1/3 . Then we have x + y + z = 0 , which implies x + y = − z .
Cubing Both Sides Cube both sides of the equation x + y = − z to get ( x + y ) 3 = ( − z ) 3 , which expands to x 3 + 3 x 2 y + 3 x y 2 + y 3 = − z 3 .
Rearranging the Equation Rearrange the equation to get x 3 + y 3 + z 3 = − 3 x 2 y − 3 x y 2 = − 3 x y ( x + y ) .
Substituting Again Substitute x + y = − z into the equation to get x 3 + y 3 + z 3 = − 3 x y ( − z ) = 3 x yz .
Substituting Back Substitute back x = a 1/3 , y = b 1/3 , and z = c 1/3 to get a + b + c = 3 ( a 1/3 ) ( b 1/3 ) ( c 1/3 ) = 3 ( ab c ) 1/3 .
Cubing Both Sides Again Cube both sides of the equation a + b + c = 3 ( ab c ) 1/3 to get ( a + b + c ) 3 = ( 3 ( ab c ) 1/3 ) 3 = 27 ab c .
Conclusion Therefore, the correct options are ( a + b + c ) 3 = 27 ab c and a + b + c = 3 ( ab c ) 1/3 .
Examples
This mathematical problem can be applied in various fields such as engineering and physics, where relationships between variables are crucial. For instance, in thermodynamics, if we consider a system where the sum of the cube roots of certain energy components is zero, this result helps simplify complex equations. Specifically, it allows us to relate the sum of the energy components to their product, providing a more manageable form for analysis and calculations. This simplification can be invaluable in designing efficient systems or understanding energy distribution.
