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

Answered by Rosalind S. Maths tutor

5445 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

How do I apply the four operations to Fractions?


How do you solve a quadratic equation? eg: x^2 + 2x - 8


How to factorise the expression x^2 + 8x + 15


Make 'a' the subject of the formula: p = (3a + 5) / (4 - a)


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences