Question

x3+1/x3+2 factorise it

Answer 1

a^3+b^3+c^3 -3.a.b.c=(a+b+c)(a^2+b^2+c^2-a.b-bc-ca)………….(1)

x^3 + 1/x^3 - 2

=x^3+1/x^3+1 -3

=(x)^3+(1/x)^3 +(1)^3–3.x.1/x.1

=(x+1/x+1)(x^2+1/x^2+1^2-x.1/x-1/x.1–1.x)

=(x+1/x+1)(x^2+1/x^2+1–1–1/x-x)

=(x+1/x+1)(x^2+1/x^2–1/x-x)

Answer 2

The factorisation of $x^3+\dfrac{1}{x^3} +2$ is$[(x+\dfrac{1}{x})+1][(+\dfrac{1}{x})-1][(x+\dfrac{1}{x})-2]$.

Step-by-step explanation:

We have,

$x^3+\dfrac{1}{x^3} +2$                             ........ (1)

To find, the factorisation of $x^3+\dfrac{1}{x^3} +2=?$

We know that,

$(x+\dfrac{1}{x})^3=x^{3}+\dfrac{1}{x^3}+3.x.\dfrac{1}{x}(x+\dfrac{1}{x})$

⇒ $(x+\dfrac{1}{x})^3=x^{3}+\dfrac{1}{x^3}+3(x+\dfrac{1}{x})$

⇒ $x^{3}+\dfrac{1}{x^3}=(x+\dfrac{1}{x})^3-3(x+\dfrac{1}{x})$  ........ (2)

From (1) and (2),

$x^3+\dfrac{1}{x^3} +2$

$=(x+\dfrac{1}{x})^3-3(x+\dfrac{1}{x})+2$

Let $x+\dfrac{1}{x}=a$

$=a^3-3a+2$$=a^3-2a-a+2$

$=a^{2}(a-2)-1(a-2)$

$=(a^{2}-1)(a-2)$

$=(a+1)(a-1)(a-2)$

Using algebraic identity,

$a^{2}-b^{2}=(a+b)(a-b)$

Put $x+\dfrac{1}{x}=a$, we get

$=[(x+\dfrac{1}{x})+1][(+\dfrac{1}{x})-1][(x+\dfrac{1}{x})-2]$

Hence, the factorisation of $x^3+\dfrac{1}{x^3} +2$ is $[(x+\dfrac{1}{x})+1][(+\dfrac{1}{x})-1][(x+\dfrac{1}{x})-2]$.