Factorise b2-c2-2 b c-a2A- b2-cc2-aB- b-c-ab-c-a- b-c-ab-c-aD- b c-ab c-a

The correct option is C (b+c+a)(b+c a) Given: b^2+ c^2+2bc a^2 ⇒ (b+c)^2 a^2 ( using (a+b)^2) ⇒ (b+c+a)(b+c a) ( using difference of squares ) ...

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Factorisation by Grouping

Factorise b2+...

Question

Factorise $b^{2} + c^{2} + 2 b c - a^{2}$

$(b + c + a) (b + c - a)$

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$(b^{2} + c) (c^{2} - a)$

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$(b - c + a) (b + c - a)$

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$(b c + a) (b c - a)$

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Solution

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Similar questions

a^{3} + b^{3} + c^{3} - 3 a b c

If a + b + c = 0, then prove the following

(a) (b + c) (b − c) + a(a + 2b) = 0

(b) a(a² − bc) + b(b² − ca) + c(c² − ab) = 0

(d)

a, b

c

\frac{a}{c} =

a, b, c

\frac{a^{2} + a b + b^{2}}{b^{2} + b c + c^{2}} = \frac{a}{c}

\frac{a^{2} + b^{2} + c^{2}}{{(a + b + c)}^{2}} = \frac{a - b + c}{a + b + c}

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