Question

If x2 + (1/x2) = 1, then what is the value of x48 + x42 + x36 + x30 + x24 + x18 + x12 + x6 + 1?

Answer 1

Answer:

The value of given expression is 1.

Step-by-step explanation:

Using the fact:

if $x+\frac{1}{x}=\sqrt{3}$,  

then,

$x^6 = -1$

If $x^2+\frac{1}{x^2}=1$

First find: $x+\frac{1}{x}$

$(x+\frac{1}{x})^2 =x^2+\frac{1}{x^2}+2$

Then;

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

⇒$x+\frac{1}{x}=\sqrt{3}$

⇒$x^6 = -1$

We have to find the value of $x^{48} + x^{42} + x^{36}+ x^{30}+ x^{24} + x^{18} + x^{12} + x^{6}+1$:

We can write this as:

$(x^{6})^8+ (x^{6})^7 + (x^{6})^6+ (x^{6})^5+ (x^{6})^4 + (x^{6})^3+ (x^{6})^2 + x^{6}+1$:

Substitute $x^6 = -1$ then;

$(-1)^8+(-1)^7+(-1)^6+(-1)^5+(-1)^4+(-1)^3+(-1)^2+(-1)+1$

⇒$1-1+1-1+1-1+1-1+1$

Simplify:

⇒$1$

Therefore, the value of  $x^{48} + x^{42} + x^{36}+ x^{30}+ x^{24} + x^{18} + x^{12} + x^{6}+1$ is, 1.