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The factors of a3+b3+c33abc are

Answer
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Hint: From the given question we have been asked to find the factors of given expression. For answering this question we will use factorization and factor theorem. We will assume that a3+b3+c33abc be a function in a and we will use the substitution method and solve the given question using factor theorem and basic mathematical operations like addition and subtraction. So, the solution will be as follows.

Complete step by step solution:
We have been given a3+b3+c33abc.
So, as mentioned in the above we will assume that,
f(a)=a3+b3+c33abc be a function in a.
Now, we will use the substitution and we will substitute a=(b+c) in the above function. So, we get the reduced as follows.
f((b+c))=((b+c))3+b3+c3+3(b+c)bc
Now we will expand this equation using basic mathematical operations like addition and subtraction.
f((b+c))=((b+c))3+(b+c)3
f((b+c)=0
Hence, by factor theorem (a+b+c) is a factor.
Now we take the given a3+b3+c33abc function.
a3+b3+c33abc
We can rewrite this expression as follows.
a3+(b3+c3)3abc
We can rewrite the above expression as follows by adding and subtracting the term 3bc(b+c)
a3+(b+c)33abc3bc(b+c)
We already got (a+b+c) as a factor. So, we sill substitute (b+c=a) in the above expression.
So, we get the expression reduced as follows.
a3+(b+c)33abc3bc(a)
a3+(b+c)33abc+3bca
a3+(b+c)3
Now we will use the formula x3+y3=(x+y)(x2+y2xy)
So, here by comparing we get (b+c=y). So, we will substitute them in this formula. Now, we get the equation reduced as follows.
(a+b+c)(a2+b2+c2+2bcabac)
So, the factors are (a+b+c),(a2+b2+c2+2bcabac)

Note: Students should have good knowledge in the concept of factorisation. Students should be able to do calculations without any errors. Students should know the formulae like,
 x3+y3=(x+y)(x2+y2xy) to solve the question.

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