If ( a+b+c )² = 289 and a²+ b²+ c² =280 , find ab +bc +ca
4.5
2/9
9
3.5
please answer with an explanation please answer fast its urgent
Answers
Answer:
3.5
Step-by-step explanation:
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Answer:
(a+b+c)
3
=4913
Step-by-step explanation:
Given : If a^2+b^2 + c^2= 280a
2
+b
2
+c
2
=280 and ab+bc+ca=\frac{9}{2}ab+bc+ca=
2
9
To find : The value of (a+b+c)^3(a+b+c)
3
Solution :
We have given,
a^2+b^2 + c^2= 280a
2
+b
2
+c
2
=280
Expand using identity, a^2+b^2 + c^2=(a+b+c)^2-2(ab+bc+ca)a
2
+b
2
+c
2
=(a+b+c)
2
−2(ab+bc+ca)
(a+b+c)^2-2(ab+bc+ca)= 280(a+b+c)
2
−2(ab+bc+ca)=280
(a+b+c)^2-2(\frac{9}{2})= 280(a+b+c)
2
−2(
2
9
)=280
(a+b+c)^2-9= 280(a+b+c)
2
−9=280
(a+b+c)^2= 280+9(a+b+c)
2
=280+9
(a+b+c)^2= 289(a+b+c)
2
=289
Taking square root both side,
(a+b+c)= \sqrt{289}(a+b+c)=
289
(a+b+c)=17(a+b+c)=17
Taking cube both side,
(a+b+c)^3=(17)^3(a+b+c)
3
=(17)
3
(a+b+c)^3=4913(a+b+c)
3
=4913
Therefore, (a+b+c)^3=4913(a+b+c)
3
=4913