Question

Evaluate using suitable identity (0.7*0.7*0.7+0.3*0.3*0.3)/(0.7*0.7+0.3*0.3-0.7*0.3) class 9

Answer 1

Answer:

please mark me as brainleast

Answer 2

The correct answer is 2.525.

Given:

(0.7*0.7*0.7+0.3*0.3*0.3)/(0.7*0.7+0.3*0.3-0.7*0.3)

To Find:

The solution of the given expression using a suitable identity.

Solution:

To solve this problem, we will be using the following identities:

1) a³+b³ = (a+b)³ - 3a²b-3ab² = (a+b)³ - 3ab(a+b)

2) a²+2ab+b² = (a+b)²

Let us solve the given expression. On simplifying it we obtain:

$\frac{0.7*0.7*0.7 &\ + &\ 0.3*0.3*0.3 }{0.7*0.7 &\ + &\ 0.3*0.3 &\ + &\ 0.7*0.3}$ = $\frac{0.7^{3} &\ + &\ 0.3^{3} }{0.7^{2} &\ + &\ 0.3^{2} &\ + &\ 0.7*0.3 }$     ..............................(A)

Let us first simplify the numerator using the identity (1). Taking a = 0.7 and b = 0.3, we get:

0.7³ + 0.3³ = (0.7+0.3)³ - 3(0.7x0.3)(0.7+0.3)

⇒ 0.7³ + 0.3³ = 1³ - 3(0.21)(0.1) = 1 - 0.063.

Hence, 0.7³ + 0.3³ = 1 - 0.063                       ..............................(B)

Next, let us simplify the denominator. Adding and subtracting 3(0.7x0.3) in the denominator, we get:

${0.7^{2} &\ + &\ 0.3^{2} &\ - &\ 0.7*0.3 }$ = 0.7² + 0.3² - 0.7*0.3 + 3(0.7*0.3) - 3(0.7*0.3)

⇒ ${0.7^{2} &\ + &\ 0.3^{2} &\ - &\ 0.7*0.3 }$ = 0.7² + 0.3² + 2(0.7*0.3) - 3(0.7*0.3)

Take a = 0.7 and b = 0.3. Using the identity in (2), we get:

${0.7^{2} &\ + &\ 0.3^{2} &\ - &\ 0.7*0.3 }$ = (0.7 + 0.3)² - 3(0.7*0.3) = 1² - 0.63

⇒ 0.7² 0.3² + 0.7*0.3 = 1 - 0.63.                     ..............................(C)

Substituting the values of (B) and (C) into (A), we get

$\frac{0.7*0.7*0.7 &\ + &\ 0.3*0.3*0.3 }{0.7*0.7 &\ + &\ 0.3*0.3 &\ + &\ 0.7*0.3}$ = (1-0.063)/(1-0.63) = 0.937/0.37 = 2.525

Hence, the correct answer is 2.525.

#SPJ2