a-maths!!!!!

α + β = - p , αβ = q

i) α²+β² = ( α + β )^2 - 2αβ

= ( - p )^2 - 2q

= p^2 - 2q

ii ) α³+β³ = ( α + β )( α² - αβ+β² )

= ( - p )( p^2 - 2q - q )

= - p ( p^2 - 3q )

iii) (α²-β-1)(β²-α-1)

= α²β² - α³ -α² - β³ + αβ +β - β² + α + 1

= α²β² - (α³+β³) - (α²+β² ) + ( α+β ) + 1 + αβ

= q^2 + p ( p^2 - 3q ) - ( p^2 - 2q ) -p + q + 1

= q^2 + p^3 - 3pq - p^2 + 2q - p + q + 1

= q^2 + p^3 - p^2 - 3q ( p - 1 ) - ( p - 1 )

= q^2 + p^2 ( p - 1 ) - ( p - 1 ) - 3 ( p - 1 )q

= q^2 + ( p^2 - 1 )(p - 1 ) - 3( p - 1 )q

= q^2 - 3 ( p - 1 )q + ( p + 1 )( p - 1 )^2

Sum of root(s): α+β= -p
Product of root(s): αβ= q


(i)α²+β²,
α²+β²= (α+β)² - 2αβ
= (-p)² -2q
= p-2q

(ii)α³+β³,
α³+β³ = (α+β)(α² -αβ+β²)
= (-p)(α²+β²-αβ)
= (-p)( p-2q-q)
= (-p)(p-3q)
= -p²+3pq

(iii)(α²-β-1)(β²-α-1)= α²β² - α³ -α² -β³ +αβ+β -β²+α+1
= (αβ)² - (α³ +β³) -(α² +β²)+αβ +(α+β)+1
= q² - (-p²+3pq)-(p-2q)+q -p+1
= q² +p²-3pq-p+2q+q -p+1
= q² +p²-3pq+3q -2p+1

α+β=-p
αβ=q

i) α²+β²=α²+β²+2αβ-2αβ
=(α+β)²-2αβ
=p²-2q

ii)α³+β³=(α+β)(α²-αβ+β²)
=-p[(α+β)²-3αβ]
=-p³+3pq

iii) (α²-β-1)(β²-α-1)=α²β²-α³-α²-β³+αβ+β-β²+α+1
=q²-(α³+β³)-(α²+β²)+αβ+α+β+1
=q²+p³-3pq-p²+2q+q-p+1

consider α+β=-p, αβ=q

(i)α²+β²=(α+β)²-2αβ
=p²-2q

(ii)α³+β³=(α+β)³-3αβ(α+β)
=-p³ -3q(-p)
=-p³ +3pq

(iii)(α²-β-1)(β²-α-1)
=α²β²-α³-α²-β³+αβ+β-β²+α+1
=(αβ)² -(α³+β³)+αβ+(α+β)-(α²+β²)+1
=q² +p³ -3pq+q-p-p²+2q+1
=q² +3(1-p)q+p³ -p²-p+1
=q² +3(1-p)q+p²(p-1)-(p-1)
=q² +3(1-p)q+(p-1)(p²-1)
=q² +3(1-p)q+(p-1)²(p+1)

兩根的和 = α+β = -p
兩根的積 = αβ = q

(i) α²+β²
= α²+2αβ+β²-2αβ
= (α+β)²-2αβ
= (-p)²-2q
= p²-2q

(ii) α³+β³
= α³+3α²β+3αβ²+β³-3α²β-3αβ²
= (α+β)³-3αβ(α+β)
= (-p)³-3q(-p)
= -p³+3pq

(iii) (α²-β-1)(β²-α-1)
= α²β²-α³-α²-β³+αβ+β-β²+α+1
= (αβ)²-(α³+β³)-(α²+β²)+(α+β)+αβ+1
= q²-(-p³+3pq)-(p²-2q)+(-p)+q+1
= q²+p³-3pq-p²+2q-p+q+1
= p³-p²+q²-p+3q-3pq+1