If a + b + c = 6, then find the value of
(2-a) + (2-6) + (2-c)² - 3(2-a) (2-6) (2-c)
Answers
Answer:
If a+b+c=6 then find the value (2-a) ^3+(2-b) ^3+(2-c) ^3-3(2-a) (2-b) (2-c)?
a+b+c=6
Therefore (2-a)+(2-b)+(2-c)=0
Now (2-a)^3+(2-b)^3+(2-c)^3–3(2-a)(2-b)(2-c)
={(2-a)+(2-b)+(2-c)}{(2-a)^2+(2-b)^2+(2-c)^2-(2-a)(2-b)-(2-b)(2-c)-(2-c)(2-a)} (since a^3+b^3+c^3–3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
={2-a+2-b+2-c}{(2-a)^2+(2-b)^2+(2-c)^2–(2-a)(2-b)-(2-b)(2-c)-(2-c)(2-a)}
={6-(a+b+c)}{(2-a)^2+(2-b)^2+(2-c)^2-(2-a)(2-b)-(2-b)(2-c)-(2-c)(2-a)}
={6–6}{(2-a)^2+(2-b)^2+(2-c)^2–(2-a)(2-b)-(2-b)(2-c)-(2-c)(2-a)}
=0×{(2-a)^2+(2-b)^2+(2-c)^2-(2-a)(2-b)-(2-b)(2-c)-(2-c)(2-a)}
=0
If (a-1)^2+(b-2)^2+(c+3)^2=0, then what is the value of a+b+c?
If a+b+c=6, a^2+b^2+c^2=12, a^3+b^3+c^3=24 , then what is the value of a-b+c?
If a+b+c=1 , a2+b2+c2=2 , a3+b3+c3=3 , how would you find a−1+b−1+c−1 ?
If a2+b2+c2+3=2(a+b+c) then what is the value of a+b+c?
If a3+b3=2 , a2+b2=2 , then what is the value of a+b ?
Zero. Let x = 2-a, y = 2-b, z = 2-c. Then a+b+c = 6 → x + y + z = 0. Your right hand expression becomes x^3 + y^3 + z^3 - 3xyz. This can be factored as
(x+y+z)*(x^2 -xy + y^2 - yz - xz + z^2), which is obviously zero since we said x+y+z was zero.
If a+b+c=1 , a2+b2+c2=2 , a3+b3+c3=3 , how would you find a−1+b−1+c−1 ?
Hi,
you can try to break your head over this by finding an analytical approach, but why not use brute force and write a program?
We can infer that a, b and c all should be in between [-sqrt(2), + sqrt(2)]
This is since a^2 + b^2 + c^2 = 2.
All squares are positive, so if any of the terms would be bigger than 2, the others can never compensate for that, since they cant be negative.
You can solve c as 1 - a - b;
Now if you substitute this in a^2 + b^2 + c^2 = 2, you get:
2*b^2 +(2*a-2)*b + (2*a^2 - 2*a -1) = 0
This is a quadratic equation in b.
Let A = 2
B = (2*a-2)
C= (2*a^2 - 2*a -1)
We find D^2 = -12*a^
If a+b+c=1 , a2+b2+c2=2 , and a3+b3+c3=3 , then what is a×b×c ?
If a3+b3=2 , a2+b2=2 , then what is the value of a+b ?
This is a problem that can be solved by a method called: BY INSPECTION. Taking the second given, and realizing that both a and b cannot be less than one (else the sum of their squares will come up less than unity), we can infer that both a and b must be unity. Let’s check: the cube of one is one, and that satisfies the first given in the question.
If BY inspection is not your cup of tea, here’s an alternate solution:
Factor the first binomial; and you get: (a + b) (a^2 - a b + b^2)
We are given that a^2 + b^2 = 2
Rewrite the factor as: (a + b) (2 - ab).
We want the product to be (2); given.
The only values of a and b that satisfy this are: a = b = 1
CHECK: (1 + 1) (2 - 1 x 1) = 2.
If (a-1)^2+(b-2)^2+(c+3)^2=0, then what is the value of a+b+c?
If a+b+c=6, a^2+b^2+c^2=12, a^3+b^3+c^3=24 , then what is the value of a-b+c?
If a+b+c=1 , a2+b2+c2=2 , a3+b3+c3=3 , how would you find a−1+b−1+c−1 ?
If a2+b2+c2+3=2(a+b+c) then what is the value of a+b+c?
If a3+b3=2 , a2+b2=2 , then what is the value of a+b ?
If a+b+c=1 , a2+b2+c2=2 , and a3+b3+c3=3 , then what is a×b×c ?
If a3+b3+c3=547 , then what is the value of a+b+c, and (a3+b3+c3)−(a2+b2+c2)−(a+b+c) ?
If a+b+c=6 and 1/a+1/b+1/c=3/2, then what is the value of a/b+a/c+b/c+b/a+c/a+c/b?
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If a+b=3 and a-b=2, then what is the value of 2 (a^2+b^2)?
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If (a^2+b^2+c^2) +3=2 (a+b+c) ^3 then what is the value of (a+b+c)?
If a-b=3 and b-c =5, then what is the value of a^2+b^2+c^2-ab-bc-CA?
How do I simplify: (a+b+c)3−(a−b−c)3−6(b+c)(a2−(b+c)2) ?