Question

class 9 maths........​

Answer 1

Question :

i) If (a + b)³ = a³ + b³ + 3ab ( a + b ), then find the value of (104)³

ii) If p(x) = y² - y + 1 , then find the value of p(1)

Answer :

i)

$\sf ({104})^{3} = (100 + 4)^{3}$

a = 100

b = 4

$\implies\sf (a + b)^3 = a^3 + b^3 + 3ab ( a + b )$

$\implies\sf (100 + 4 )^3 = (100)^3 + 4^3 + 3 \times 100 \times 4 ( 100 + 4 )$

$\implies\sf (104)^3 = 1000000 + 64 + 1200 ( 104 )$

$\implies\sf (104)^3 = 1000064 + 124800$

$\implies\sf\boxed{ (104)^3 = 1124864}$

ii)

$\sf p(x) = y^2 - y + 1$

To find p(1) , we need to substitute the value of y = 1 in the equation.

$\implies\sf p(1) = 1^2 - 1 + 1$

$\implies\sf p(1) = \cancel 1 - \cancel 1 + 1$

$\implies\boxed{\sf p(1) = 1}$

Answer 2

Step-by-step explanation:

$Question :</p><p></p><p>i) If (a + b)³ = a³ + b³ + 3ab ( a + b ), then find the value of (104)³</p><p></p><p>ii) If p(x) = y² - y + 1 , then find the value of p(1)</p><p></p><p>Answer :</p><p></p><p>i)</p><p></p><p>\sf ({104})^{3} = (100 + 4)^{3}(104)3=(100+4)3</p><p></p><p>a = 100</p><p></p><p>b = 4</p><p></p><p>\implies\sf (a + b)^3 = a^3 + b^3 + 3ab ( a + b )⟹(a+b)3=a3+b3+3ab(a+b)</p><p></p><p>\implies\sf (100 + 4 )^3 = (100)^3 + 4^3 + 3 \times 100 \times 4 ( 100 + 4 )⟹(100+4)3=(100)3+43+3×100×4(100+4)</p><p></p><p>\implies\sf (104)^3 = 1000000 + 64 + 1200 ( 104 )⟹(104)3=1000000+64+1200(104)</p><p></p><p>\implies\sf (104)^3 = 1000064 + 124800⟹(104)3=1000064+124800</p><p></p><p>\implies\sf\boxed{ (104)^3 = 1124864}⟹(104)3=1124864</p><p></p><p>ii)</p><p></p><p>\sf p(x) = y^2 - y + 1p(x)=y2−y+1</p><p></p><p>To find p(1) , we need to substitute the value of y = 1 in the equation.</p><p></p><p>\implies\sf p(1) = 1^2 - 1 + 1⟹p(1)=12−1+1</p><p></p><p>\implies\sf p(1) = \cancel 1 - \cancel 1 + 1⟹p(1)=1−1+1</p><p></p><p>\implies\boxed{\sf p(1) = 1}⟹p(1)=1</p><p></p><p>$