Question

simplify: (ii) (x2+x+1) (x2
-x+1) and find its value for x=1

plz give me answer​

Answer 1

Step-by-step explanation:

(x²+x+1)(x²-x+1) = x⁴-x³+x²+x³-x²+x+x²-x+1 = x⁴+x²+1

if x=1. x⁴+x²+1= 1+1+1= 3

Answer 2

Given,

$(x^{2} +x+1)$×$(x^{2} -x+1)$.

To Find,

The simplified value of this.

Also, the value of this is when x=1.

Solution,

We just need to calculate it in the way as follows,

$(x^{2} +x+1)$×$(x^{2} -x+1)$

=$[(x^{2} +1)+x]$×$[(x^{2} +1)-x]$.

[There's a rule that if (a+b)×(a-b)=$a^{2} -b^{2}$]

[Here a =$(x^{2} +1)$ and b = x.]

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

=[$(x^{2} )^{2} +2.x^{2} .1+1 -x^{2}$]

=$x^{4} +x^{2} +1$.

Now when x= 1 then the value  $x^{4} +x^{2} +1$ will be, $1^{4} +1^{2}+1$=3.

Hence, The simplified value is  $x^{4} +x^{2} +1$ , and when the value of x=1 then the value of  $x^{4} +x^{2} +1$  is 3.