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

6391 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

The equation the line L1 is y=3x-2 and the equation of line L2 is 3y-9x+5=0. Show that these two lines are parallel.


Factorise x^2+12x+7


Solve the simultaneous equations x + y = 2 and x^2 + 2y = 12


The point P has coordinates (3, 4) The point Q has coordinates (a, b) A line perpendicular to PQ is given by the equation 3x + 2y = 7 Find an expression for b in terms of a.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning