Question

What is the value of \(\frac{7^3+2^3+84}{7^2-2^2}\) ?

(a) 9 \(\frac{1}{2}\)

(b) 9 \(\frac{2}{3}\)

(c) 9 \(\frac{1}{3}\)

(d) 9 \(\frac{1}{4}\)

I had been asked this question in class test.

I'd like to ask this question from Binomial Theorem for Positive Integral Index topic in division Binomial Theorem of Mathematics – Class 11

Answer 1

Right option is (b) 9 \(\frac{2}{3}\)

Best explanation: Using binomial theorem we know that (a + b)^3 = a^3 + 3ab^2 + 3a^2b + b^3

Therefore, (7 + 2)^3 = 7^3 + 2^3 + (3 x 7 x 2^2) + (3 x 2 x 7^2)

(9)^3 = 7^3 + 2^3 + 84 + (3 x 2 x 7^2)

729 = 7^3 + 2^3 + 84 + 294

7^3 + 2^3 + 84 = 435

Also 7^2 – 2^2 = (7 – 2)(7 + 2)

7^2 – 2^2 = (5)(9)

7^2 – 2^2 = 45

So \(\frac{7^3 + 2^3 + 84}{7^2-2^2}\) = 435 / 45

= 9 \(\frac{30}{45}\)   

= 9 \(\frac{2}{3}\).