Question

Using the identity and proof: $(x-y{)}^4=x^4-4x^{3}y+6x^2{y}^2-4x{y}^3+y^4$.

Answer 1

$(x-y{)}^4=x^4-4x^{3}y+6x^2{y}^2-4x{y}^3+y^4$
First take L.H.S $(x-y{)}^4$
So, the above expression is written as $\left(\right(x-y{)}^2{)}^2$
We know that $(x-y{)}^2=x^2-2xy+y^2,(x-y+z{)}^2=x^2+y^2+z^2-2xy-2yz+2zx$
Therefore, $(x^2-2xy+y^2{)}^2$
Here, x = x^2, y = 2xy, z = y^2
So, $(x^2-2xy+y^2{)}^2=x^4+4x^2{y}^2+y^4-4x^{3}y-4x{y}^3+2y^2{x}^2$
$=$ $x^4-4x^{3}y+6x^2{y}^2-4x{y}^3+y^4$
L.H.S $=$ R.H.S
$x^4-4x^{3}y+6x^2{y}^2-4x{y}^3+y^4=x^4-4x^{3}y+6x^2{y}^2-4x{y}^3+y^4$
Hence, $(x-y{)}^4=x^4-4x^{3}y+6x^2{y}^2-4x{y}^3+y^4$ is proved.