Factorise: a3 -8b3 -64c3 -24abc -Maths 9th

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Factorise: a3 -8b3 -64c3 -24abc -Maths 9th

asked Feb 3, 2022 by anonymous

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Factorisation

answered Feb 3, 2022 by anonymous

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Factorise: a3 -8b3 -64c3 -24abc -Maths 9th

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If a + b + c =0, then a3+b3 + c3 is equal to -Maths 9th

Description : If a + b + c =0, then a3+b3 + c3 is equal to -Maths 9th

Last Answer : (d) Now, a3+b3 + c3 = (a+ b + c) (a2 + b2 + c2 – ab – be – ca) + 3abc [using identity, a3+b3 + c3 – 3 abc = (a + b + c)(a2+b2+c2 –ab–bc-ca)] = 0 + 3abc [∴ a + b + c = 0] a3+b3 + c3 = 3abc.

If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3 – 3abc = -25. -Maths 9th

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If a + b + c =0, then a3+b3 + c3 is equal to -Maths 9th

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Last Answer : (d) Now, a3+b3 + c3 = (a+ b + c) (a2 + b2 + c2 – ab – be – ca) + 3abc [using identity, a3+b3 + c3 – 3 abc = (a + b + c)(a2+b2+c2 –ab–bc-ca)] = 0 + 3abc [∴ a + b + c = 0] a3+b3 + c3 = 3abc.

If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3 – 3abc = -25. -Maths 9th

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The product (a + b) (a – b) (a2 – ab + b2) (a2 + ab + b2) is equal to (a) a6 + b6 (b) a6 – b6 (c) a3 – b3 (d) a3 + b3 -Maths 9th

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Last Answer : a+b+c=9 and a2+b2+c2=35 Using formula, (a+b+c)2=a2+b2+c2+2(ab+bc+ca) 92=35+2(ab+bc+ca) 2(ab+bc+ca)=81−35=46 (ab+bc+ca)=23 using formula, (a3+b3+c3)−3abc=(a2+b2+c2−ab−bc−ca)(a+b+c) a3+b3+c3−3abc=(35−23)×9=9×12=108

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Last Answer : Given, a−b=4,ab=21 Then, ⇒(a−b)3=a3−3a2b+3ab2−b3=a3−3ab(a−b)−b3 ⇒43=a3−b3−3(21)(4) ⇒64+252=a3−b3 ∴a3−b3=316

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