Question

Solve the equation :- a³-a = 0

Answer 1

Given ,

=> a³-a

=> a(a²-1) = 0

=> a(a+1)(a-1)= 0

=> either a = 0 or (a+1)(a-1) = 0

=> now a = 0 , - 1 , 1 .

Therefore solutions for a; a = 0 , -1 , +1

Answer 2

Hey there !!

Here we shall factorize
${a}^{3} - a \: = 0$
we can write it as

$a( {a}^{2} - 1) = 0 \: \: \: \: \: \: \: ( \: taking \: common)$
then
we know a formula
${a}^{2} - {b}^{2} = (a + b)(a - b)$
applying it to get ,
$a(a + 1)(a - 1) = 0$

now we can find out its root
a = 0
a +1 = > a =-1
a -1 => a = 1
u can verify it by putting this value

Hope this helps u !!