Question

factorise
a⁴- b⁴
p⁴-81
x⁴-(y+z)⁴
x⁴-(x-z)⁴

​

Answer 1

Answer:

i)a

4

−b

4

=(a

2

−b

2

)(a

2

+b

2

)=(a+b)(a−b)(a

2

+b

2

)

=(a+b)(a

2

+b

2

)(a−b)

ii)p

4

−81

=(p

2

−9)(p

2

+9)

=(p+3)(p−3)(p

2

+9)

iii)x

4

−(y+z)

4

=(x

2

−(y+z)

2

)(x

2

+(y+z)

2

)

=(x+y+z)(x−y−z)(x

2

+(y+z)

2

)

iv)x

4

−(x−z)

4

=(x

2

−(x−z)

2

)(x

2

+(x−z)

2

)

=(x+x−z)(x−x+z)(x

2

+(x−z)

2

)

=(2x−z)(z)(x

2

+(x−z)

2

)

v)a

4

−2a

2

b

2

+b

4

=a

4

−a

2

b

2

−a

2

b

2

+b

4

=a

2

(a

2

−b

2

)−b

2

(a

2

−b

2

)

=(a

2

−b

2

)

2

=(a+b)

2

(a−b)

2

Step-by-step explanation:

this is the answer bro.

hope it's help you.

Answer 2

Factorise

i) a⁴- b⁴

ii)p⁴-81

iii)x⁴-(y+z)⁴

iv)x⁴-(x-z)⁴

step by step explanation

i

$\sf=a⁴-b⁴$

$\sf=(a²)²-(b²)²$

$\sf=(a²-b²)(a²+b²)$

$\sf=(a+b)(a-3)(a²+b²)$

ii

$\sf=p⁴-3⁴$

$\sf=(p²)²-(3²)²$

$\sf=(p²-3²)(p²+3²)$

$\sf=(p+3)(p-3)(p²+9)$

iii

$\sf{ = x⁴-(y+z)⁴ }$

$\sf{ = x⁴-(y+z)⁴ }$

$\sf{ = (x²)²-((y+z)²)² }$

$\sf{ = (x²-(y+z)²)(x²+(y+z)²) }$

$\sf{ = (x²-(y+z)²)(x²+(y+z)²) }$

$\sf{ = (x+y+z)(x²+y²+z²+2yz ) }$

iv

$\sf{= x⁴-(x-z)⁴ }$

$\sf{= (x²)²-((x-z)²)² }$

$\sf{= (x²-(x-z)²)(x²+(x-z)²) }$

$\sf{= (x+x-z)(x-x+z)(x²+x²+z²-2xz) }$

$\sf{= (2x-z)(z)(2x²+z²-2xz) }$

$\sf{= (2xz-z²)(2x²+z²-2xz) }$

$\\ \\ \sf{For\ more\ information }$

$\\ \\ \sf(a + b)^{2} = {a}^{2} + {b}^{2} + 2ab$

$\sf(a - b)^{2} = {a}^{2} + {b}^{2} - 2ab$

$\sf(a + b)(a - b) = {a}^{2} - {b}^{2}$

$\sf(a + b + c)^{2} = {a}^{2} + {b}^{2} + {c}^{2}+ 2ab + 2bc + 2ca$

$\sf(a + b) ^{3} = {a}^{3} + b^{3} + 3ab(a + b)$

$\sf(a - b) ^{3} = {a}^{3} - b^{3} - 3ab(a - b)$

$\sf a ^{3} + {b}^{3} = (a + b)(a ^{2} + {b}^{2} - ab)$

$\sf a ^{3} - {b}^{3} = (a - b)(a ^{2} + {b}^{2} + ab)$