Question

Find the value of ${(a^2+\sqrt{a^2-1})}^4+{(a^2-\sqrt{a^2-1})}^4.$

Answer 1

Given,

${(a^2+\sqrt{a^2-1})}^4+{(a^2-\sqrt{a^2-1})}^4$

${(a^2+\sqrt{a^2-1})}^4:a^8+4a^6\sqrt{a^2-1}+6a^6-5a^4+4a^2{(a^2-1)}^{\frac{3}{2}}-2a^2+1$

${(a^2+\sqrt{a^2-1})}^4:a^8+4a^6\sqrt{a^2-1}+6a^6-5a^4+4a^2{(a^2-1)}^{\frac{3}{2}}-2a^2+1$

$=a^8+4a^6\sqrt{a^2-1}+6a^6-5a^4+4a^2{(a^2-1)}^{\frac{3}{2}}-2a^2+1+a^8+4a^6\sqrt{a^2-1}+6a^6-5a^4+4a^2{(a^2-1)}^{\frac{3}{2}}-2a^2+1$

$=2a^8+12a^6-10a^4-4a^2+2$