Question

If u can do these 3 correctly 33,34,35 then I’ll mark u as Brainliest

Answer 1

Question - 33

Evaluate the following :

$\sf \: \dfrac{ {a}^{2} - {(b - c)}^{2} }{ {(a + b)}^{2} - {c}^{2} } + \dfrac{ {b}^{2} - {(c - a)}^{2} }{ {(b + c)}^{2} - {a}^{2} } + \dfrac{ {c}^{2} - {(a + b)}^{2} }{ {(c + a)}^{2} - {b}^{2} }$

Consider,

$\sf \: \dfrac{ {a}^{2} - {(b - c)}^{2} }{ {(a + b)}^{2} - {c}^{2} } + \dfrac{ {b}^{2} - {(c - a)}^{2} }{ {(b + c)}^{2} - {a}^{2} } + \dfrac{ {c}^{2} - {(a + b)}^{2} }{ {(c + a)}^{2} - {b}^{2} }$

We know

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

So, using this, we get

$\sf=\dfrac{(a + b - c)(a - b + c)}{(a + b - c)(a + b + c)} + \\ \: \:\sf \: \dfrac{(b + c - a)(b - c + a)}{(b + c - a)(b + c + a)} + \\ \sf \: \dfrac{(c + a - b)(c - a + b)}{(c + a - b)(c + a + b)}$

$\sf \:  =  \: \dfrac{a - b + c}{a + b + c} + \dfrac{b - c + a}{a + b + c} + \dfrac{c - a + b}{a + b + c}$

$\sf \:  =  \: \dfrac{\cancel{a} - \cancel{b} + c + \cancel{b} - \cancel{c} + a + \cancel{c} - \cancel{a} + b}{a + b + c}$

$\sf \:  =  \: \dfrac{a + b + c}{a + b + c}$

$\sf \:  =  \: 1$

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Question- 34

Evaluate the following :-

$\sf \: \dfrac{ {a}^{2} - {b}^{2} }{ {a}^{3} - {b}^{3} } \times \dfrac{ {a}^{3} + {a}^{2}b + {ab}^{2} }{ {a}^{2} + 2ab + {b}^{2} } \times (a + b)$

Consider,

$\rm :\longmapsto\: \sf \: \dfrac{ {a}^{2} - {b}^{2} }{ {a}^{3} - {b}^{3} } \times \dfrac{ {a}^{3} + {a}^{2}b + {ab}^{2} }{ {a}^{2} + 2ab + {b}^{2} } \times (a + b)$

We know that,

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

$\boxed{ \tt{ \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2}) \: }}$

$\boxed{ \tt{ \: {x}^{2} + 2xy + {y}^{2} = {(x + y)}^{2} \: }}$

So, using these Identities, we get

$\sf= \dfrac{(a + b)(a - b)}{(a - b)( {a}^{2} + ab + {b}^{2})} \times \dfrac{a( {a}^{2} + ab + {b}^{2})}{ {(a + b)}^{2} } \times (a + b)$

$\rm \:  =  \: a$

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Question :- 35

Evaluate the following :-

$\sf \: \dfrac{ {b}^{2} + bc + {c}^{2} }{(a - b)(c - a)} + \dfrac{ {c}^{2} + ca + {a}^{2} }{(b - c)(a - b)} + \dfrac{ {a}^{2} + ab + {b}^{2} }{(c - a)(b - c)}$

Consider

$\sf \: \dfrac{ {b}^{2} + bc + {c}^{2} }{(a - b)(c - a)} + \dfrac{ {c}^{2} + ca + {a}^{2} }{(b - c)(a - b)} + \dfrac{ {a}^{2} + ab + {b}^{2} }{(c - a)(b - c)}$

can be rewritten as

$\: \sf = \: \dfrac{(b - c)({b}^{2} + bc + {c}^{2} )}{(a - b)(c - a)(b - c)} + \\ \: \: \: \: \: \: \sf \dfrac{(c - a)({c}^{2} + ca + {a}^{2} )}{(b - c)(a - b)(c - a)} + \\ \: \: \: \: \sf \: \dfrac{(a - b)({a}^{2} + ab + {b}^{2} )}{(c - a)(b - c)(a - b)}$

$\: \sf = \: \dfrac{ {b}^{3} - {c}^{3} }{(a - b)(c - a)(b - c)} + \\ \: \: \: \: \: \: \sf \dfrac{ {c}^{3} - {a}^{3} }{(b - c)(a - b)(c - a)} + \\ \: \: \: \: \sf \: \dfrac{ {a}^{3} - {b}^{3} }{(c - a)(b - c)(a - b)}$

$\sf \:  =  \: \dfrac{ {b}^{3} - {c}^{3} + {c}^{3} - {a}^{3} + {a}^{3} - {b}^{3} }{(a - b)(b - c)(c - a)}$

$\sf \:  =  \: \dfrac{ 0 }{(a - b)(b - c)(c - a)}$

$\sf \:  =  \: 0$

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More Identities to know :-

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)