How do you solve #root3(y-7)=4#?

Question

1667 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-13T08:06:46

#y=71#

raising both terms to the cube

#(y-7)=4^3#

#4^3=64#

adding up 7

#(y)=64+7#

#(y)=71#

Answer 2 · 2017-11-13T08:09:48

#y=71#

#root(3)(y-7)=4#

or, #(root(3)(y-7))^3=4^3#

or, #y-7=64#

or, #y=64+7=71#

So the value of #y# is #71#.