find the three zeroes of the polynomial p(x) = x3 – 64X – 14,
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Answered by
0
make it brllint
Step-by-step explanation:
α3+β3+γ3=(α+β+γ)(α2+β2+γ2−γβ−βγ−γα)+3αβγ
∴P(x)=x3−64x−14
∴α+β+γ=a−b=10=0
αβ+βγ+γα=ac=−64
∴αβγ=a−d=14
∴β3+β3+γ3=(0)(α2+β2+γ2−αβ−βγ−γα)+3(14)
∴α3+β3+γ3=42
Answered by
0
Answer:
Step-by-step explanation:
α
3
+β
3
+γ
3
=(α+β+γ)(α
2
+β
2
+γ
2
−γβ−βγ−γα)+3αβγ
∴P(x)=x
3
−64x−14
∴α+β+γ=
a
−b
=
1
0
=0
αβ+βγ+γα=
a
c
=−64
∴αβγ=
a
−d
=14
∴β
3
+β
3
+γ
3
=(0)(α
2
+β
2
+γ
2
−αβ−βγ−γα)+3(14)
∴α
3
+β
3
+γ
3
=42
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