Question

x^3 -1/x^3 - 14 factorise

Answer 1

Hi friend...
Here is your answer

x3−1x3−14x3−1x3−14

Now, the only way to factorise this in a non-radical, non-complex way is to use the identity-

a3+b3+c3−3abc=a3+b3+c3−3abc=

(a+b+c)(a2+b2+c2−ab−bc−ca)(a+b+c)(a2+b2+c2−ab−bc−ca)

We have only 2 cubed terms though

((a3=x3a3=x3 and b3=b3=−1x3)−1x3).

The c3−3abcc3−3abc has been merged into −14−14. So, put-

x3+−1x3−14x3+−1x3−14

≡x3+−1x3+c3−3⋅x⋅−1x⋅c≡x3+−1x3+c3−3⋅x⋅−1x⋅c

==x3+−1x3+c3+3⋅cx3+−1x3+c3+3⋅c

Comparing the constants we have-

−14=c3+3c−14=c3+3c

Testing all rational roots- c=−2c=−2 is a root.

Now we know,

a=xa=x

b=−1xb=−1x

c=−2c=−2

So, the factorisation of our expression is the RHSRHS of the identity.

Plugging in the values and simplifying, we get-

(x−1x−2)(x2+1x2+2x−2x+5)

I hope it will help
#yahyaahmad#

Answer 2

i too am suffering to solve this questions thanks though lol