Answer:
Explanations:
According to the question, we need to determine which of the signs will fit in that will make the expression a binomial.
In simple terms, a binomial is a two-term algebraic expression that contains variable, coefficient, exponents, and constant.
We need to determine the required sign by using the trial and error method.
Using the positive sign (+) first, we will have:
[tex]\begin{gathered} =\mleft(3y^6+4\mright)+(9y^{12}-12y^6+16) \\ =3y^6+4+9y^{12}-12y^6+16 \\ =3y^6-12y^6+4+9y^{12}+16 \\ =-9y^6+9y^{12}+20 \end{gathered}[/tex]
Using the product sign, this will be expressed as:
[tex]\begin{gathered} (3y^6+4)\cdot(9y^{12}-12y^6+16) \\ (3y^6+4)\cdot\lbrack(3y^6)^2-(3y^6)(4)^{}+4^2)\rbrack \end{gathered}[/tex]
According to the sum of two cubes;
[tex]a^3+b^3=\mleft(a+b\mright)•(a^2-ab+b^2)[/tex]
Comparing this with the expression above, we will see that a = 3y^6 and
b = 4. This means that the resulting expression above can be written as a sum of two cubes to have;
[tex]\begin{gathered} (3y^6+4)\cdot\lbrack(3y^6)^2-(3y^6)(4)^{}+4^2)\rbrack^{} \\ =(3y^6)^3-4(3y^6)^2+4(3y^6)^2+16(3y^6)+4(3y^6)^2-16(3y^6)+4^3 \\ \end{gathered}[/tex]
Collect the like terms:
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