what whould be the formula of [a-(3b+1)[a+(3b+1)] using the special product?

Question

what i mean is using the formula's of special product, what formula fit in the problem like this [a-(3b+1)][a+(3b+1)]

Answer 1

[a - (3b + 1)][ a + (3b + 1)] = a² - (3b + 1)²
and this may be the answer you are after.
If desired, the brackets may be expanded to give:-
a² - (9b² + 6b + 1) = a² - 9b² - 6b - 1

Answer 2

What is this "special product", may I know? Anyway, the breakup of your problem goes like this :-

[a-(3b+1)][a+(3b+1)] = a^2 - (3b+2)^2
= a^2 - {(3b)^2 + 2*3b*2 + 4}
= a^2 - 9b^2 - 12b - 4

Answer 3

Using the product formula, the answer is
a^2 - (3b+1)^2
which when expanded equals,
a^2-9b^2-6b-1

Answer 4

Statements
[a-(3b+1)] [a+(3b+1)]
(a)^2 - (3b+1)^2 Difference of Squares
a^2 - 9b^2 + 6b + 1

is that it? not sure what you meant by "special product"

Answer 5

You can use this formula ....( a - b) (a +b) = a^2 - b^2

[ a - (3b +1)][ a + (3b+1)]
a^2 - ( 3b +1)^2

a^2 - ( 9b^2 + 6b + 1)
a^2- 9b^2 - 6b -1......this is my ans.

Answer 6

take 3b+1 as x
this is of the form [a-b][a+b]=a^2-b^2
theyfore {a-(3b+1)] [a+(3b+1)]=a^2-(3b+1)^2
=a^2-(9b^2+6b+1)
=a^2-9b^2-6b-1

Answer 7

[a-(3b+1)][a+(3b+1)]=
=a^2-(3b+1)^2

Answer 8

-a(3b-1)^2=
(a-b)^2=a^2-2ab +b^2