If a and b are positive consecutive even integers, where a > b, which of the
following is equal to a3-b3?
A) 6b2 +12b-8
B)6b2 - 12b - 8
C)6b2+12b + 8
D)6b2-12b + 8
E)None of these
Answers
Answer:
We have
a and b are two odd positive integers such that a & b
but we know that odd numbers are in the form of 2n+1 and 2n+3 where n is integer.
so, a=2n+3, b=2n+1, n∈1
Given ⇒ a>b
now, According to given question
Case I:
2
a+b
=
2
2n+3+2n+1
=
2
4n+4
=2n+2=2(n+1)
put let m=2n+1 then,
2
a+b
=2m ⇒ even number.
Case II:
2
a−b
=
2
2n+3−2n−1
2
2
=1 ⇒ odd number.
Hence we can see that, one is odd and other is even.
This is required solutions.
Answer:
The correct answer is option(c)
Step-by-step explanation:
Given,
a and b are consecutive positive even integers and a>b
To find,
The value of a³-b³
Recall the formula
(x+y)³ = x³ + 3x²y + 3xy² + y³ ---------------(1)
Solution:
Since a and b are two consecutive positive even numbers and a>b,
we can take
a = b+2
a³-b³ = (b+2)³ - b³ ---------------(2)
applying the identity (1) with x = b and y = 2 we get
(b+2)³ = b³ + 3×b²×2 + 3×b×2² + 2³
=b³ + 6b² + 12b + 8
Substituting (2) we get
a³-b³ = b³ + 6b² + 12b + 8- b³
= 6b² + 12b + 8
∴ a³-b³ = 6b² + 12b + 8
Answer:
a³-b³ = 6b² + 12b + 8
The correct answer is option(c)
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