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If a+b+c=11 a2+b2+c2=75 then a3+b3+c3-3abc = ? give full explanation pls |
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Answer» 572Step-by-step explanation:a+b+c=11---(1)a²+b²+c²=75---(2)(2) BECOMES(a+b+c)²-2(ab+bc+ca)=75(11)²-2(ab+bc+ca)=75121-2(ab+bc+ca)=75-2(ab+bc+ca)=75-121-2(ab+bc+ca)=-46- get cancel both side2(ab+bc+ca)=46(ab+bc+ca)=46/2(ab+bc+ca)=23---(3)a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)=(a+b+c){(a²+b²+c²)-(ab+bc+ca)}by substituting (1)(2)(3)=11{75-23}=11(52)=572 |
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