Question

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Answer 1

Question 1

(103)³

Solution:-

(103)³ = (100+3)³

We know that,

(a+b)³ = a³+b³+3ab(a+b)

using this identity,

→(100+3)³ = (100)³+(3)³+3×100×3(100+3)

→(100+3)³ = 1,000,000+27+900(103)

→ (103)³ = 1,000,027+92,700

→(103)³=1,092,727

Answer:- 1,092,727.

Question 2

(99)²

(99)² = (100-1)²

Identity:-

(a-b)²=a²+b²-2ab

Solution:-

→(100-1)² = (100)² + (1)² - 2 × 100×1

→(100-1)² = 10,000+ 1 - 200

→(100-1)² = 10,001-200

→(100-1)² = 10,001-200

→(100-1)²=9,801.

Answer:- (99)² = 9,801.

QUESTION 3:-

$\rm \dfrac{9}{16}x^2-\dfrac{1}{9}y^2$

$\rm\implies\Bigg ( \dfrac{3}{4}x \Bigg )^2-\Bigg (\dfrac{1}{3}y\Bigg )^2$

Identity:-

a²-b² = (a+b)(a-b)

Solution:-

$\rm\implies\Bigg ( \dfrac{3}{4}x \Bigg )^2-\Bigg (\dfrac{1}{3}y\Bigg )^2$

$\rm\implies\Bigg ( \dfrac{3}{4}x + \dfrac{1}{3}y \Bigg )\Bigg ( \dfrac{3}{4}x -\dfrac{1}{3}y \Bigg ) \qquad\qquad ...[since, \ a^2-b^2 = (a+b)(a-b)]$

$\rm\implies\Bigg ( \dfrac{9x+4y}{12} \Bigg )\Bigg ( \dfrac{9x-4y}{12} \Bigg )$

$\rm\implies \dfrac{9x+4y}{12} \times </p><p>\dfrac{9x-4y}{12}$

$\rm\implies \dfrac{(9x+4y)(9x-4y)}{12}$

$\rm\implies \dfrac{(9x)^2-(4y)^2}{12} \qquad\qquad ...[since , \ (a+b)(a-b)=a^2-b^2]$

$\rm\implies \dfrac{81x^2-16y^2}{12}$