Question

plz answer the following question
${x}^{3} - (1 \div {x}^{3} ) - 14$
It is not 1 ÷ x cube
It is 1/x cube
Plz answer

Answer 1

hey buddy,

x^3−1/x^3−14

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

a^3+b^3+c^3−3abc=

(a+b+c)(a^2+b^2+c^2−ab−bc−ca)

We have only 2 cubed terms though

(a^3=x^3 and b^3=−1/x^3)

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

x3+ −1/x^3−14

≡x^3+ −1/x^3 + c^3 −3* X* −1/x*c

=x^3+ −1x^3 + c^3+ 3*c

Comparing the constants we have-

−14 = c^3 +3c

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

Now we know,

a = x

b= −1/x

c= −2

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

Plugging in the values and simplifying, we get-

(x −1/x −2)
(x^2 + 1/x^2 + 2x −2/x+ 5)