Question

prove that
$(x + y) ^{3} - (x - y) ^{3} - 6y( {x}^{2} - {y}^{2} ) = {8y}^{3}$

Answer 1

${(x + y)}^{3} - {(x - y)}^{3} - 6y( {x}^{2} - {y}^{2}) = 8 {y}^{3}$
Expand by using formula

$( {x}^{3} + 3 {x}^{2} y + 3x {y}^{2} + {y}^{3} ) - ( {x}^{3} - 3 {x}^{2} y + 3x {y}^{2} - {y}^{3} ) - 6 {x}^{2} y + 6 {y}^{3} = 8 {y}^{3}$
Remove the brackets

${x}^{3} + 3 {x}^{2} y + 3x {y}^{2} + {y}^{3} - {x}^{3} + 3 {x}^{2} y - 3x {y}^{2} + {y}^{3} - 6 {x}^{2}y + 6 {y}^{2} = 8 {y}^{3}$
Eliminate the opposites

$0 + {y}^{3} + {y}^{3} + 6 {y}^{3} = 8 {y}^{3}$

Add similar terms

$8 {y}^{3} = 8 {y}^{3}$




L.H.S = R.H.S

HENCE PROVED