ax-acube -bx+bcube=0 findx
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Answered by
2
Hi ,
ax - a³ - bx + b³ = 0
ax - bx - ( a³ - b³ ) = 0
x( a - b ) - ( a - b ) ( a² + ab + b² ) = 0
( a - b ) [ x - ( a² + ab + b² ) ] = 0
Therefore ,
a - b = 0 or x - ( a² + ab + b² ) = 0
x = a² + ab + b²
I hope this helps you
: )
ax - a³ - bx + b³ = 0
ax - bx - ( a³ - b³ ) = 0
x( a - b ) - ( a - b ) ( a² + ab + b² ) = 0
( a - b ) [ x - ( a² + ab + b² ) ] = 0
Therefore ,
a - b = 0 or x - ( a² + ab + b² ) = 0
x = a² + ab + b²
I hope this helps you
: )
prabhupada1:
thanks
Answered by
1
ax-a^3-bx+b^3 = 0
x(a-b)-(a^3-b^3) = 0
x(a-b)-(a-b)(a^2+b^2+ab) =0
x = a^2+b^2+ab
x(a-b)-(a^3-b^3) = 0
x(a-b)-(a-b)(a^2+b^2+ab) =0
x = a^2+b^2+ab
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