Lols Question (maths) (1 Viewer)

[super.muppy]

Factorise x^6-1 first as (x^3)^2-1, then go on
similarly factor: x^8 - a^8, x^10-1

and one more

by expressing x as (root x)^2 and y as (root y)^2, factor

(x^3/2)-(y^3/2)


show working plz thanks

 

[kurt.physics]

Factor x^6-1 first as (x^3)^2-1, then go on
similarly factor: 1 x^8 - a^8, 2 x^10-1

and one more

by expressing x as (root x)^2 and y as (root y)^2, factor

(x^3/2)-(y^3/2)


show working plz thanks

1 x⁸ - a⁸ = (x⁴)² - (a⁴)²
=(x⁴ - a⁴)(x⁴ + a⁴)
=((x²)² - (a²)²)(x⁴ + a⁴)
=(x² - a²)(x² + a²)(x⁴ + a⁴)
=(x - a)(x + a)(x² + a²)(x⁴ + a⁴)

 

[suzlee]

factor = factorise?

I didn't get what the question was at first... lol.

 

[RachelElizabeth]

Factor x^6-1 first as (x^3)^2-1, then go on
similarly factor: x^8 - a^8, x^10-1

and one more

by expressing x as (root x)^2 and y as (root y)^2, factor

(x^3/2)-(y^3/2)


show working plz thanks

I am sooo not ready for that.. I have no hope!

 

[kurt.physics]

I'll do the other ones in a few minutes

 

[super.muppy]

I'll do the other ones in a few minutes

???

 

[kurt.physics]

1 Factorise x^10-1

and one more

2 by expressing x as (root x)^2 and y as (root y)^2, factor

(x^3/2)-(y^3/2)

show working plz thanks

1 x¹⁰ - 1
= (x⁵)² - 1²
= (x⁵ - 1)(x⁵ + 1)

And thats as far as it goes

For the second one, i presume you mean this

(x^3/2)-(y^3/2) is x^3/2 - y^3/2

In which case, substituting ( (root x)² and similarly with y) we get

= ((root x)²)^3/2 - ((root y)²)^3/2

= (root x)^{2 * 3/2} - (root y)^{2 * 3/2}

= (root x)³ - (root y)³

Now we apply the difference of two cubes ( ie a² - b² = (a - b)(a² + ab + b²) )

so,

x^3/2 - y^3/2 = (root(x) - root(y))(x +root(xy) + y)


P.S. Sorry about the lateness :S