What is the factored form of t6 − p3? A.
(t2 + p)(t4 − pt2 + p2)
B.
(t − p)2(t2 − pt + p2)
C.
(t2 − p)(t4 + pt2 + p2)
D.
(t − p)2(t2 + pt + p2)

(C). (t² - p)( + pt² + p²)

Step-by-step explanation:

a³ - b³ = (a - b)(a² + ab + b²)

- p³ = (t²)³ - p³ = (t² - p)( + pt² + p²)

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°