[(a+b+c)⁴-(b+c)⁴-(a+c)⁴-(a+b)⁴+a⁴+b⁴+c⁴]/(a+b+c)

asked by guest
on Oct 07, 2024 at 7:50 pm



You asked:

Evaluate the expression: \(\frac{{\left( a + b + c \right)}^{4} - {\left( b + c \right)}^{4} - {\left( a + c \right)}^{4} - {\left( a + b \right)}^{4} + {a}^{4} + {b}^{4} + {c}^{4}}{a + b + c}\)

MathBot Answer:

Evaluated



\(\displaystyle \frac{{\left( a + b + c \right)}^{4} - {\left( b + c \right)}^{4} - {\left( a + c \right)}^{4} - {\left( a + b \right)}^{4} + {a}^{4} + {b}^{4} + {c}^{4}}{a + b + c} = \frac{a^{4} + b^{4} + c^{4} - \left(a + b\right)^{4} - \left(a + c\right)^{4} - \left(b + c\right)^{4} + \left(a + b + c\right)^{4}}{a + b + c} \)

Expanded

\[\frac{{\left( a + b + c \right)}^{4} - {\left( b + c \right)}^{4} - {\left( a + c \right)}^{4} - {\left( a + b \right)}^{4} + {a}^{4} + {b}^{4} + {c}^{4}}{a + b + c} = \frac{12 a^{2} b c}{a + b + c} + \frac{12 b^{2} a c}{a + b + c} + \frac{12 c^{2} a b}{a + b + c}\]

Factored

\[\frac{{\left( a + b + c \right)}^{4} - {\left( b + c \right)}^{4} - {\left( a + c \right)}^{4} - {\left( a + b \right)}^{4} + {a}^{4} + {b}^{4} + {c}^{4}}{a + b + c} = 12 a b c\]

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