Show that:(i) 11+xa-b+11+xb-a=1(ii) xaa-bxaa+b÷xbb-axbb+aa+b=1(iii) x1a-b1a-cx1b-c1b-ax1c-a1c-b=1(iv) xa2+b2xaba+bxb2+c2xbcb+cxc2+a2xaca+c=x2a3+b3+c3(v) xa-ba+bxb-cb+cxc-ac+a=1(vi) xa-a-11a-1aa+1=x(vii) ax+1ay+1x+yay+2az+2y+zaz+3ax+3z+x=1(viii) 3a3ba+b3b3cb+c3c3ac+a=1

Show that:(i) 11+xa-b+11+xb-a=1(ii) xaa-bxaa+b÷xbb-axbb+aa+b=1(iii) x

1.	Show that:(i) 11+xa-b+11+xb-a=1(ii) xaa-bxaa+b÷xbb-axbb+aa+b=1(iii) x1a-b1a-cx1b-c1b-ax1c-a1c-b=1(iv) xa2+b2xaba+bxb2+c2xbcb+cxc2+a2xaca+c=x2a3+b3+c3(v) xa-ba+bxb-cb+cxc-ac+a=1(vi) xa-a-11a-1aa+1=x(vii) ax+1ay+1x+yay+2az+2y+zaz+3ax+3z+x=1(viii) 3a3ba+b3b3cb+c3c3ac+a=1
Answer» Show that: (i) $\frac{1}{1 + x^{a - b}} + \frac{1}{1 + x^{b - a}} = 1$ (ii) ${[\{\frac{x^{a (a - b)}}{x^{a (a + b)}}\} \div \{\frac{x^{b (b - a)}}{x^{b (b + a)}}\}]}^{a + b} = 1$ (iii) ${(x^{\frac{1}{a - b}})}^{\frac{1}{a - c}} {(x^{\frac{1}{b - c}})}^{\frac{1}{b - a}} {(x^{\frac{1}{c - a}})}^{\frac{1}{c - b}} = 1$ (iv) ${(\frac{x^{a^{2} + b^{2}}}{x^{a b}})}^{a + b} {(\frac{x^{b^{2} + c^{2}}}{x^{b c}})}^{b + c} {(\frac{x^{c^{2} + a^{2}}}{x^{a c}})}^{a + c} = x^{2 (a^{3} + b^{3} + c^{3})}$ (v) ${(x^{a - b})}^{a + b} {(x^{b - c})}^{b + c} {(x^{c - a})}^{c + a} = 1$ (vi) ${\{{(x^{a - a^{- 1}})}^{\frac{1}{a - 1}}\}}^{\frac{a}{a + 1}} = x$ (vii) ${(\frac{a^{x + 1}}{a^{y + 1}})}^{x + y} {(\frac{a^{y + 2}}{a^{z + 2}})}^{y + z} {(\frac{a^{z + 3}}{a^{x + 3}})}^{z + x} = 1$ (viii) ${(\frac{3^{a}}{3^{b}})}^{a + b} {(\frac{3^{b}}{3^{c}})}^{b + c} {(\frac{3^{c}}{3^{a}})}^{c + a} = 1$