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

5965 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Convert 0.1727272... to a fraction in its lowest terms.


Complete the square and find the roots (where y=0) of: "x^2+4x-5"


Complete the square for x^2 + x - 6.


How would I simplify this? (x-2)(x+3)


We're here to help

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

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences