Which of the equations are true identities? A. (x^(2)+2y)(x^(2)-3y)=x^(4)-6y^(2) B. k^(3)-r^(3)=(k^(2)+r)(k-r^(2)) Choose 1 answer: (A) Only A (B) Only B (C) Both A and B (D) Neither A nor B

Expand using FOIL method: Check equation A: (x^(2)+2y)(x^(2)-3y) Use the FOIL method to expand. x^(2) * x^(2) - x^(2) * 3y + 2y * x^(2) - 2y * 3ySimplify the terms: Simplify the terms. x^(4) - 3x^(2)y + 2x^(2)y - 6y^(2)Combine like terms: Combine like terms. x^(4) - x^(2)y - 6y^(2) This is not equal to x^(4) - 6y^(2), so equation A is not a true identity.Apply difference of cubes formula: Check equation B: k^(3)-r^(3)=(k^(2)+r)(k-r^(2)) Use the difference of cubes formula: a^3 - b^3 = (a - b)(a^2 + ab + b^2)Use formula on k^(3)-r^(3): Apply the formula to k^(3)-r^(3). (k - r)(k^2 + kr + r^2) This is not equal to (k^(2)+r)(k-r^(2)), so equation B is not a true identity.

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Which of the equations are true identities?
A.
(x^(2)+2y)(x^(2)-3y)=x^(4)-6y^(2)
B.
k^(3)-r^(3)=(k^(2)+r)(k-r^(2))
Choose 1 answer:
(A) Only A
(B) Only B
(C) Both A and B
(D) Neither A nor B

Which of the equations are true identities? $\newline$ A. $\left(x^{2}+2 y\right)\left(x^{2}-3 y\right)=x^{4}-6 y^{2}$ $\newline$ B. $k^{3}-r^{3}=\left(k^{2}+r\right)\left(k-r^{2}\right)$ $\newline$ Choose $1$ answer: $\newline$ (A) Only A $\newline$ (B) Only B $\newline$ (C) Both A and B $\newline$ (D) Neither A nor B

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Q. Which of the equations are true identities?

\newline

\left(x^{2}+2 y\right)\left(x^{2}-3 y\right)=x^{4}-6 y^{2}

\newline

k^{3}-r^{3}=\left(k^{2}+r\right)\left(k-r^{2}\right)

\newline

Choose

1

answer:

\newline

(A) Only A

\newline

(B) Only B

\newline

\newline

(D) Neither A nor B

Expand using FOIL method: Check equation A: $x^{2}+2y)(x^{2}-3y)\ Use the FOIL method to expand. \$x^{2}$ $*$ $x^{2}$ - $x^{2}$ $*$ $3y$ + $2y$ $*$ $x^{2}$ - $2y$ $*$ $3y$
Simplify the terms: Simplify the terms. $x^{4} - 3x^{2}y + 2x^{2}y - 6y^{2}$
Combine like terms: Combine like terms. $\newline$ $x^{4} - x^{2}y - 6y^{2}$ $\newline$ This is not equal to $x^{4} - 6y^{2}$ , so equation A is not a true identity.
Apply difference of cubes formula: Check equation B: $k^{3}-r^{3}=(k^{2}+r)(k-r^{2})$ $\newline$ Use the difference of cubes formula: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$
Use formula on $k^{3}-r^{3}$ : Apply the formula to $k^{3}-r^{3}$ . $\newline$ $(k - r)(k^2 + kr + r^2)$ $\newline$ This is not equal to $(k^{2}+r)(k-r^{2})$ , so equation B is not a true identity.