If$$\ x^{2}$$- 3x + 1 = 0, then the value of $$\ x^{5}+\frac{1}{x^{5}}\ $$is equal to

$$x^{2}-3x+1$$=0

Taking 'x' common

x(x-3+$$\frac{1}{x})$$=0

$$\Rightarrow x+\frac{1}{x}$$=$$3\rightarrow(1)$$

Squaring on both sides

$$x^{2}+\frac{1}{x^{2}}+2\times x\times\frac{1}{x}$$=9

$$\Rightarrow x^{2}+\frac{1}{x^{2}}$$=$$7\rightarrow(2)$$

Cubing equation(1) on both sides

$$x^{3}+\frac{1}{x^{3}}+3\times x\times\frac{1}{x}(x+\frac{1}{x})$$=27

$$x^{3}+\frac{1}{x^{3}}$$+$$3\times 1\times3$$=27($$\because x+\frac{1}{x}$$=3)

$$x^{3}+\frac{1}{x^{3}}$$=27-9=$$18\rightarrow(3)$$

Squaring equation(2) on both sides

$$x^{4}+\frac{1}{x^{4}}+2\times x^{2}\times\frac{1}{x^{2}}$$=49

$$x^{4}+\frac{1}{x^{4}}$$=$$47\rightarrow(4)$$

Multiplying equation(1) and equation(4)

$$(x^{4}+\frac{1}{x^{4}})(x+\frac{1}{x}$$)=$$47\times3$$

$$x^{5}+\frac{1}{x^{5}}+x^{3}+\frac{1}{x^{3}}$$=$$47\times3$$=141

$$x^{5}+\frac{1}{x^{5}}+18$$=141($$\because x^{3}+\frac{1}{x^{3}}$$)

$$\therefore x^{5}+\frac{1}{x^{5}}$$=123