a+b+c =8, ab+bc+ca = 20 find the value of a^3+b^3+ c^3- 3abc
Don't spam
Attachments:
Answers
Answered by
1
Answer:
Step-by-step explanation:
Since (a+b+c)
2
=a
2
+b
2
+c
2
+2ab+2bc+2ca
(a+b+c)
2
=a
2
+b
2
+c
2
+2(ab+bc+ca)
a+b+c=8 and ab+bc+ca=20,
(8)
2
=a
2
+b
2
+c
2
+2×(20)
⇒64=a
2
+b
2
+c
2
+40
∴a
2
+b
2
+c
2
=64−40=24
We know that
a
3
+b
3
+c
3
−3abc
=(a+b+c){a
2
+b
2
+c
2
−(ab+bc+ca)}
∴a
3
+b
3
+c
3
−3abc
=8×(24−20)=4×8=32
[∵a+b+c=8,ab+bc+ca=20 and a
2
+b
2
+c
2
=24]
Thus, a
3
+b
3
+c
3
−3abc=32
Answered by
0
menishray13 please i-n-b-o-x me
Similar questions