Rs aggarwal solutions
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Factorization of Polynomials | Factorization of Polynomials Exercise 3F

Question 37

(0.85×0.85×0.85+0.15×0.15×0.15)/(0.85×0.85−0.85×0.15+0.15×0.15)= 1

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(0.85×0.85×0.85+0.15×0.15×0.15)/(0.85×0.85−0.85×0.15+0.15×0.15)= 1

Let us take 0.85 = a and 0.15 = b

So we know that LHS

=(𝑎×𝑎×𝑎+𝑏×𝑏×𝑏)/(𝑎×𝑎−𝑎×𝑏+𝑏×𝑏)

=(𝑎^3+𝑏^3)/(𝑎^2−𝑎𝑏+𝑏^2)

According to the equation,

𝑎^3 + 𝑏^3 = (𝑎 + 𝑏)(𝑎^2 − 𝑎𝑏 + 𝑏^2)

So we get,

=[(𝑎+𝑏)(𝑎^2−𝑎𝑏+𝑏^2)]/(𝑎^2−𝑎𝑏+𝑏^2)

Since (𝑎^2 − 𝑎𝑏 + 𝑏^2) is similar in numerator and denominator

We get,

= (a + b)

Now by replacing the values of a and b

= 0.85 + 0.15

= 1

= RHS

Hence proved

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Our top 5% students will be awarded a special scholarship to Lido.

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Question 37

(0.85×0.85×0.85+0.15×0.15×0.15)/(0.85×0.85−0.85×0.15+0.15×0.15)= 1

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

(0.85×0.85×0.85+0.15×0.15×0.15)/(0.85×0.85−0.85×0.15+0.15×0.15)= 1

Let us take 0.85 = a and 0.15 = b

So we know that LHS

=(𝑎×𝑎×𝑎+𝑏×𝑏×𝑏)/(𝑎×𝑎−𝑎×𝑏+𝑏×𝑏)

=(𝑎^3+𝑏^3)/(𝑎^2−𝑎𝑏+𝑏^2)

According to the equation,

𝑎^3 + 𝑏^3 = (𝑎 + 𝑏)(𝑎^2 − 𝑎𝑏 + 𝑏^2)

So we get,

=[(𝑎+𝑏)(𝑎^2−𝑎𝑏+𝑏^2)]/(𝑎^2−𝑎𝑏+𝑏^2)

Since (𝑎^2 − 𝑎𝑏 + 𝑏^2) is similar in numerator and denominator

We get,

= (a + b)

Now by replacing the values of a and b

= 0.85 + 0.15

= 1

= RHS

Hence proved

Our top 5% students will be awarded a special scholarship to Lido.

subject-cta
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