square root of 98 + cube root of 32 - sixth root of 50.
Hello, varungk92!
Are you sure of the wording of the problem?
It doesn't simplify very much and it's a really stupid problem.
I suspect that it's suppose to be: .$\displaystyle \sqrt{98} + 3\sqrt{32} - 6\sqrt{50}$
You're expected to know how to simplify radicals: .$\displaystyle \sqrt{ab} \:=\:\sqrt{a}\cdot\sqrt{b}$
The first term is: .$\displaystyle \sqrt{98}\;=\;\sqrt{49\cdot2}\;=\;\sqrt{49}\cdot\s qrt{2} \;= \;7\sqrt{2}$
The second term is: .$\displaystyle 3\sqrt{32}\;=\;3\sqrt{16\cdot2}\;=\;3\sqrt{16}\cdo t\sqrt{2}\;=\;3\cdot4\cdot\sqrt{2} \;=\;12\sqrt{2}$
The third term is: .$\displaystyle 6\sqrt{50}\;=\;6\sqrt{25\cdot2}\;=\;6\cdot\sqrt{25 }\cdot\sqrt{2}\;=\;6\cdot5\cdot\sqrt{2}\;=\;30\sqr t{2}$
Then the problem becomes: .$\displaystyle 7\sqrt{2} + 12\sqrt{2} - 30\sqrt{2}$
Answer: .$\displaystyle \boxed{-11\sqrt{2}}$