Question

how to solve help me there are 3 question .......................... . . ..​

Answer 1

Answer:

Step-by-step explanation:

1)

Given :

$a^x = c^q = b$

&

$c^y = a^z = d$

Then,.

Solution :

$a^x = c^q = b$  (given)

⇒ $a^x = b$

⇒ $a = b^{\frac{1}{x}}$  ...(i)

⇒  $c^q = b$

⇒ $c = b^{\frac{1}{q}}$ ..(ii)

$c^y = a^z = d$

By substituting ,

$c = b^{\frac{1}{q}}$ & $a = b^{\frac{1}{x}}$

⇒ $c^y = a^z = d$

⇒ $(b^{\frac{1}{q}})^{y} = (b^{\frac{1}{x}})^{z} = d$

⇒ $b^{\frac{y}{q}} = b^{\frac{z}{x}}$

⇒ $\frac{y}{q}= \frac{z}{x}$

⇒ $xy = qz$

______

2)

Given :

To find the value of :

(a + b)³ + (a - b)³ + 6a(a² - b²)

Solution :

we know that,

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

⇒ (a + b)³ + (a - b)³ + 6a(a² - b²)

⇒ [a³ + 3a²b + 3ab² + b³] - [a³ - 3a²b + 3ab² - b³] + 6a³ - 6ab²

⇒ [ 2a³ + 6ab² ] + 6a³ - 6ab²

⇒ 8a³

_____

3)

Given :

To factorize :

x⁴ + 5x³ + 5x² - 5x - 6

Solution :

⇒ x⁴ + 5x³ + 5x² - 5x - 6

⇒ x⁴ + 2x³ + 3x³ + 6x² - x² - 2x - 3x - 6

⇒ x³ (x + 2) + 3x² (x + 2) - x (x + 2) - 3 (x + 2)

⇒ (x + 2) (x³ + 3x² - x - 3)

⇒ (x + 2) ( x² (x + 3) - 1(x + 3))

⇒ (x + 2) (x² - 1) (x + 3)

(or)

⇒ (x + 2) ( x (x + 1) - 1 (x + 1) ) (x + 3)

⇒ (x + 2) (x + 1) (x - 1) (x + 3)