Question

Factorise the expression,
(x+y+z)³-x³-y³-z³

into linear factors with complete steps

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Answer 1

Answer:

Let p(x)=q(y)=r(z)= (x+y+z)³-x³-y³-z³

Now, if (x+y) (y+z) (z+x) are the factors of the polynomial, then (x+y), (y+z) & (z+x) individually are also its factors.

Now, let x+y=0

=> x=-y

Given that, p(x)=(x+y+z)³-x³-y³-z³

Then, p(-y)=(-y+y+z)³-(-y)³-y³-z³ = z³+y³-y³-z³=0

∴ by factor theorem (x+y) is a factor of p(x)

Again, let y+z=0

=> y=-z

Given that, q(y) =(x+y+z)³-x³-y³-z³

Then, q(-z) = (x-z+z)³-x³-(-z)³-z³=x³-x³+z³-z³=0

∴ by factor theorem (y+z) is a factor of q(y) [=p(x)]

Similarly, we can easily prove that (z+x) is a factor of r(z) [=q(y)=p(x)]

Step-by-step explanation:

I hope this helps u

Answer 2

Answer:

general form, we can write it as 1/x-1 + 0. Clearly, the degree of this polynomial is not one, it is not a linear polynomial.