Question

what's the answer please ​

Answer 1

Answer:

Use the formula

$\boxed{ \red{ {x}^{2} - {y}^{2} = (x + y)(x - y) }}$

(a)

$p = \frac{59 {}^{2} - 51 {}^{2} }{8} \\ \\ = \frac{(59 + 51)(59 - 51)}{8} \\ \\ = \frac{110 \times 8}{8} = 110$

(b)

$p = \frac{ {76}^{2} - 67 {}^{2} }{143} \\ \\ = \frac{(76 + 67)(76 - 67)}{143} \\ \\ = \frac{143 \times 6}{143} =6$

(c)

$p = \frac{(3a + b) {}^{2} - (3a - b) {}^{2} }{ab} \\ \\ = \frac{(3a + b + 3a - b)(3a + b - 3a + b)}{ab } \\ \\ = \frac{6a \times 2b}{ab} = \frac{12ab}{ab} = 12$

(d)

$441 - {p}^{2} = {21}^{2} - {17}^{2} \\ \\ \implies \: {21}^{2} - {p}^{2} = {21}^{2} - {17}^{2} \\ \\ \implies {p}^{2} - {17}^{2} + 21^{2} - {21}^{2} = 0 \\ \\ \implies \: (p + 17)(p - 17) = 0 \\ \\ \implies \: p = 17 \ \: \: \: or\: \: - 17$

Answer 2

Answer:

$\qquad\qquad\underline{\textsf{\textbf{ \color{magenta}{Solution\:completed} }}}$

Solution :-

[Please refer that attachment for your answer]

☞Some Basic Algebraic Identities-

★ (a + b)² = a² + 2ab + b²

★ (a − b)² = a² − 2ab + b²

★ (a + b) (a – b) = a² – b²

★ (x + a) (x + b) = x² + (a + b)x + ab

★ (x + a) (x – b) = x² + (a – b)x – ab

★ (x – a) (x + b) = x² + (b – a)x – ab

★ (x – a) (x – b) = x² – (a + b)x + ab

★ (a + b)³ = a³ + b³ + 3ab (a + b)

★ (a – b)³ = a³ – b³ – 3ab (a – b)

HOPE IT HELPS :)