If $$x^{2}+\frac{1}{x^{2}}=1$$, then what is the value of $$x^{48}+x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^{6}+1$$?

Given : $$x^{2}+\frac{1}{x^{2}}=1$$

Adding '2' on both sides,

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

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

=> $$x+\frac{1}{x}=\sqrt3$$ ----------(i)

Now, cubing both sides, we get :

=> $$x^3+\frac{1}{x^3}+3(x)(\frac{1}{x})(x+\frac{1}{x})=(\sqrt3)^3$$

=> $$x^3+\frac{1}{x^3}+3\sqrt3=3\sqrt3$$     [Using equation (i)]

=> $$\frac{x^6+1}{x^3}=0$$

=> $$x^6=-1$$ -----------(ii)

To find : $$x^{48}+x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^{6}+1$$

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

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

= $$(1)+(-1)+(1)+(-1)+(1)+(-1)+(1)+(-1)+(1)$$

= $$1$$

=> Ans - (C)