Question & Answer
QUESTION

Find the factors of ${\left( {a + b + c} \right)^4} - {\left( {b + c} \right)^4} - {\left( {c + a} \right)^4} - {\left( {a + b} \right)^4} + {a^4} + {b^4} + {c^4}.$

ANSWER Verified Verified

Hint: In these types of problems, remember that the factor of an expression is the value at which the value of expression becomes zero. Analyze the given expression and find out the answer.


Complete step by step answer:


We are given with the expression,

$\Rightarrow {\left( {a + b + c} \right)^4} - {\left( {b + c} \right)^4} - {\left( {c + a} \right)^4} - {\left( {a + b} \right)^4} + {a^4} + {b^4} + {c^4}{\text{ (Eq 1)}}$

Now, for finding the factors of expression at equation 1,

Let put b = 0  in the equation. It becomes,

$\Rightarrow {\left( {a + 0 + c} \right)^4} - {\left( {0 + c} \right)^4} - {\left( {c + a} \right)^4} - {\left( {a + 0} \right)^4} + {a^4} + {0^4} + {c^4}$

On solving, above expression,

$\Rightarrow {\left( {a + c} \right)^4} - {\left( c \right)^4} - {\left( {c + a} \right)^4} - {\left( a \right)^4} + {a^4} + {c^4} = 0$

So, as we see that value of expression at equation 1 becomes 0 when b = 0.

Let’s put a = 0 in the equation 1. It becomes,

$\Rightarrow {\left( {0 + b + c} \right)^4} - {\left( {b + c} \right)^4} - {\left( {c + 0} \right)^4} - {\left( {0 + b} \right)^4} + {0^4} + {b^4} + {c^4}$

On solving, above expression,

$\Rightarrow {\left( {b + c} \right)^4} - {\left( {b + c} \right)^4} - {\left( c \right)^4} - {\left( b \right)^4} + {b^4} + {c^4} = 0$

So, as we see that value of expression at equation 1 becomes 0 when a = 0.

Let’s put c = 0 in the equation 1, It becomes,

$\Rightarrow {\left( {a + b + 0} \right)^4} - {\left( {b + 0} \right)^4} - {\left( {0 + a} \right)^4} - {\left( {a + b} \right)^4} + {a^4} + {b^4} + {0^4}$

On solving, above expression,

$\Rightarrow {\left( {a + b} \right)^4} - {\left( b \right)^4} - {\left( a \right)^4} - {\left( {a + b} \right)^4} + {a^4} + {b^4} = 0$


So, as we see that value of expression at equation 1 becomes 0 when c = 0. 

Now , putting a + b + c = 0, c + a = - b,a + b = - c and b + c = - a

$\Rightarrow {\left( 0 \right)^4} - {\left( { - a} \right)^4} - {\left( { - b} \right)^4} - {\left( { - c} \right)^4} + {a^4} + {b^4} + {c^4}$

On solving, above expression ,

$\Rightarrow - {\left( a \right)^4} - {\left( b \right)^4} - {\left( c \right)^4} + {a^4} + {b^4} + {c^4} = 0$

So, as we see that value of expression at equation 1 becomes 0 when a + b + c = 0 

As equation 1 is a fourth power equation. And value of it becomes 0 at a = 0,b = 0,c = 0 and a + b + c = 0.

So, we can write equation 1 as,

$\Rightarrow {\left( {a + b + c} \right)^4} - {\left( {b + c} \right)^4} - {\left( {c + a} \right)^4} - {\left( {a + b} \right)^4} + {a^4} + {b^4} + {c^4}{\text{ }} = kabc(a + b + c){\text{ (Eq 2)}}$

where k is constant in equation 2


So, for finding the value of k, we put a = 1,b = 2,c = 3 in equation 2 we get,

$\Rightarrow {\left( {1 + 2 + 3} \right)^4} - {\left( {2 + 3} \right)^4} - {\left( {3 + 1} \right)^4} - {\left( {1 + 2} \right)^4} + {1^4} + {2^4} + {3^4}{\text{ }} = k*1*2*3*(1 + 2 + 3)$

On solving, above expression , it becomes

$\Rightarrow 1296 - 625 - 256 - 81 + 1 + 16 + 81 = 36k$

On simplification we get,

$\Rightarrow$ 432 = 36k

$\Rightarrow$ So, k = 12 

Now, putting value of k in equation 2 we get,

$\Rightarrow {\left( {a + b + c} \right)^4} - {\left( {b + c} \right)^4} - {\left( {c + a} \right)^4} - {\left( {a + b} \right)^4} + {a^4} + {b^4} + {c^4}{\text{ }} = 12abc(a + b + c)$

Hence factors of the given expression at equation 1 are (a = 0),(b = 0),(c = 0) and (a + b + c = 0).


NOTE: - Whenever you come up with this type of problem then always remember that the factor of an expression is the value at which value of expression becomes zero.