[tex]\displaystyle\bf\\\Big(b-3\Big)\Big(b^{2} +3b+9\Big)-b\Big(b-4\Big)\Big(b+4\Big)=0\\\\\\b^{3} -3^{3} -b\cdot\Big(b^{2} -16\Big)=0\\\\\\b^{3} -27-b^{3} +16b=0\\\\\\16b-27=0\\\\\\16b=27\\\\\\b=1\frac{11}{16}[/tex]
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[tex]\displaystyle\bf\\\Big(b-3\Big)\Big(b^{2} +3b+9\Big)-b\Big(b-4\Big)\Big(b+4\Big)=0\\\\\\b^{3} -3^{3} -b\cdot\Big(b^{2} -16\Big)=0\\\\\\b^{3} -27-b^{3} +16b=0\\\\\\16b-27=0\\\\\\16b=27\\\\\\b=1\frac{11}{16}[/tex]