how am I meant to solve sq.root(6^2+8^2) = cube.root(125a^3) when one side is squared and the other is cubed?

first of all, have a look at each side separately to see you you can cancel anything down.e.g. the square and square root can cancel out. => sq.root(62+82) = 6+8on the otherside, the cube, and cube root can cancel out => cube.root (125a3) -> cube.root(a3) = a, leaving (cube.root(125))a (is 125 a cube number?)125 is 53 leaving an equation with no squares, cubes or roots on either sidesq.root(62+82) = cube.root(125a3) goes too => 6+8=5a=> 6+8=15 therefore 15=5a=> a=15/5=> a=3

RS
Answered by Rosalind S. Maths tutor

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