al pure maths substitution.123

(a) a + b + c = 0 => a = -(b + c)

a^3 + b^3 + c^3 + 3abc

= -(b + c)^3 + b^3 + c^3 - 3bc(b + c)

= -(b^3 + 3bc(b + c) + c^3) + b^3 + c^3 - 3bc(b + c)

= 0

(b) Consider a = x - 2√2 + 5√3, b = -3x - 6√2 + 7√3, c = 2x + 8√2 - 12√3a + b + c

= x - 2√2 + 5√3 -3x - 6√2 + 7√3 + 2x + 8√2 - 12√3

= -2√2 + 5√3 - 6√2 + 7√3 + 8√2 - 12√3

= 0

So, the original equation become

3abc = 0

3(x - 2√2 + 5√3)(-3x - 6√2 + 7√3)(2x + 8√2 - 12√3) = 0

x = 2√2 - 5√3, (7√3 - 6√2)/3, 6√3 - 4√2

thz Kenneth

[8^(1/3) = 2]

c^3 = 8(x + 4√2 - 6√3)^3
= 2^3 x ( x + 4√2 - 6√3 )^3
= [ 2 ( x + 4√2 - 6√3 ) ] ^3 <---- law of index, a^m x b^m = (ab)^m
= (2x + 8√2 - 12√3) ^3