Question

Please please answer fast ​

Answer 1

Answer:

Formula

X²-y²

(x+y) ²-2xy

Solution

X²-y²

(x+y) ²-2xy

15²-2×54

225-108

117

Step-by-step explanation:

Hope it helps.....

Answer 2

$\blue\bigstar$Answer:

$\sf\: \boxed {\sf {x}^{2} \: - \: {y}^{2} \: = \: \tt \: </u>\<u>p</u><u>m</u><u> </u><u>45}$

$\red\bigstar$ Given:

• ( x + y ) = 15
• xy = 54

$\pink\bigstar$ To Find:

• x² - y²

$\blue\bigstar$ Solution:

We know that :

• (x + y)² = x² + 2xy + y²

By substituting the given values, we get :

$\sf \: {(15 )}^{2} \: = \: {x}^{2} \: + \: {y}^{2} \: + 2(54)$

$\sf\implies \: 225 \: = \: {x}^{2} \: + \: {y}^{2} \: + \: 108$

$\sf \implies \: 225 \: - \: 108 \: = \: {x}^{2} \: + \: {y}^{2}$

$\sf \implies \: \boxed{ \tt{x}^{2} \: + \: {y}^{2} \: = \: 117}$

We know that :

• (x - y)² = x² - 2xy + y²

So by substituting the values, we get :

$\sf { (x \: - \: y)}^{2} = {x}^{2} \: + {y}^{2} \: - 2xy$

$\sf\implies {(x \: - \: y)}^{2} \: = \: 117 \: - \: 2(54)$

$\sf\implies \: {(x \: - \: y)}^{2} \: = 9$

$\sf\implies \: (x \: - \: y) = \: \sqrt{9}$

$\sf \implies \boxed{ \sf \: (x \: - \: y) \: = \: \pm 3 }$

$\sf \boxed{ \sf \therefore \: \: (x \: - \: y) \: = \: \pm 3 }$

Now, we know that :

• x² - y² = (x + y)(x - y)

So by substituting the values, we get :

Case 1:

$\sf\implies \: {x}^{2} \: - \: {y}^{2} \: = \: \tt \: (15)(3)$

$\sf\implies \: \boxed {\sf {x}^{2} \: - \: {y}^{2} \: = \: \tt \: 45}$

$\sf\: \boxed {\sf \therefore \: {x}^{2} \: - \: {y}^{2} \: = \: \tt \: 45}$

Case 2:

$\sf\implies \: {x}^{2} \: - \: {y}^{2} \: = \: \tt \: (15)(-3)$

$\sf\implies \: \boxed {\sf {x}^{2} \: - \: {y}^{2} \: = \: \tt \: -45}$

$\sf\: \boxed {\sf \therefore \: {x}^{2} \: - \: {y}^{2} \: = \: \tt \: -45}$

$\red\bigstar$ Concepts Used:

• (x + y)² = x² + 2xy + y²
• Substitution of values
• Transposition Method
• (x - y)² = x² - 2xy + y²
• x² - y² = (x + y)(x - y)

$\green\bigstar$Extra - Information:

• (y + a)(y + b) = y² + (a + b)y + ab

• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

• (a + b)³ = a³ + b³ + 3ab (a + b)

• (a – b)³ = a³ – b³ – 3ab (a – b)

• a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)

• If (a + b + c ) = 0, then a³ + b³ + c³ = 3abc