If a + b + c = 9 and ab + bc + ca = 23, then a3 + b3 + c3 – 3 abc = (a) 108 (b) 207 (c) 669 (d) 729 -Maths 9th

If a + b + c = 9 and ab + bc + ca = 23, then a3 + b3 + c3 – 3 abc = (a) 108 (b) 207 (c) 669 (d) 729 -Maths 9th
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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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