Simple Algebra problems

$We\quad know\quad that\quad \frac { 1 }{ a } +\frac { 1 }{ b } =6\\ Multiplying\quad both\quad sides\quad with\quad ab,\quad we\quad get\\ a+b=6ab\\ But,\quad a+b=6\\ So,\quad 6=6ab\\ or,\quad ab=1\\ \\ Now,\quad squaring\quad both\quad sides\quad of\quad a+b=6\\ { a }^{ 2 }+{ b }^{ 2 }+2ab\quad =\quad 36\\ { a }^{ 2 }+{ b }^{ 2 }+2\quad =\quad 36\\ { a }^{ 2 }+{ b }^{ 2 }\quad =\quad 34\\ \\ Now\quad solving\quad the\quad given\quad equation,\\ \frac { a }{ b } +\frac { b }{ a } +1981\\ \frac { a^{ 2 }+{ b }^{ 2 } }{ ab } +1981\\ \frac { 34 }{ 1 } +1981\\ 2015$

CHEERS!:)

$\frac{1}{a}+\frac{1}{b}=\frac{b+a}{ab}=\frac{6}{ab}=6$ Dividing both sides by $6$ we get: $\frac{1}{ab}=1\;so\;ab=1$ Squaring both sides of $a+b=6$ ,we get: $(a+b)^2=6^2$ $a^2+2ab+b^2=36$ Replacing the value of $ab$ in this equation we get: $a^2+2(1)+b^2=36$ $a^2+b^2=36-2=\boxed{34}$ Now we can solve $\frac{a}{b}+\frac{b}{a}+1981$ as follows: Rearrange $\frac{a}{b}+\frac{b}{a}\;to\;get\;\frac{a^2+b^2}{ab}$ Replacing the respective values in the equation we get: $\frac{34}{1}+1981=34+1981=\boxed{2015}$

$The\quad answer\quad is:\\ "THE\quad NEXT\quad YEAR"$