question no. 73 and 74
Attachments:
Answers
Answered by
1
73. Given a + b + c = 0
Then a + b = -c ----- (1)
On cubing both sides, we get
(a+b)^3 = (-c)^3
a^3 + b^3 + 3ab(a+b) = -c^3
a^3 + b^3 + 3ab(-c) = -c^3
a^3 + b^3 - 3abc = -c^3
a^3 + b^3 + c^3 = 3abc
(a^3 + b^3 + c^3)/(abc) = 3
(a^2/bc + b^2/ca + c^2/ab) = 3.
74) Given a + b + c = 5 -------- (1)
Given ab + bc + ca = 10 ----- (2).
We know that a^3 + b^3 + c^3 - 3abc = (a + b + c)^3 - 3(a+b+c)(ab+bc+ca)
= (5)^3 - 3(5)(10)
= 125 - 150
= -25.
Therefore a^3 + b^3 + c^3 - 3abc = -25.
Hope this helps!
Then a + b = -c ----- (1)
On cubing both sides, we get
(a+b)^3 = (-c)^3
a^3 + b^3 + 3ab(a+b) = -c^3
a^3 + b^3 + 3ab(-c) = -c^3
a^3 + b^3 - 3abc = -c^3
a^3 + b^3 + c^3 = 3abc
(a^3 + b^3 + c^3)/(abc) = 3
(a^2/bc + b^2/ca + c^2/ab) = 3.
74) Given a + b + c = 5 -------- (1)
Given ab + bc + ca = 10 ----- (2).
We know that a^3 + b^3 + c^3 - 3abc = (a + b + c)^3 - 3(a+b+c)(ab+bc+ca)
= (5)^3 - 3(5)(10)
= 125 - 150
= -25.
Therefore a^3 + b^3 + c^3 - 3abc = -25.
Hope this helps!
siddhartharao77:
Sorry ur formula is correct
I thought ur asking question no 73.
a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
ya
(a+b+c)^2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
i know that one
Okkk
2 more questions please...
Now i have some work..Anyways i will try now or later
ok☺☺
Similar questions